This is one possible solving path.
The only palindromic two-digit prime is 11
(22 etc are multiples of 11, not primes),
so **23ac = 11**.

21dn is a multiple of 10 and it’s the digit sum of 29ac, which is ≤ (9 + 9 + 9), ie 21dn is one of {10, 20}.

14ac contains only odd digits, so the square 11dn ends in one of {1, 5, 9}. Then 14ac is one of {135, 531, 579, 975}; the last two have a digit product of 315, which isn’t a triangular number (the 24th and 25th triangular numbers are 300 and 325), so 14ac is restricted to {135, 531} and 11dn is one of {25, 81}.

2dn is a palindromic multiple of 5 (and isn’t allowed to start with 0), so it matches 5n5 and the digit product is 25×n. If n is odd, the digit product ends in 25 or 75 and the digit product of that has more than one digit, so it takes more than two steps to reach a single digit, ie 2dn’s multiplicative persistence isn’t 2. Thus n must be even (not 0) so that 2dn’s digit product is a multiple of 50 and the next digit product is 0, ie 2dn is one of {525, 545, 565, 585}.

5dn is 2n^{2}, ie one of {18, 32, 50, 72, 98}.
We can eliminate 50 because 10ac mustn’t start with 0.
3ac is a square, which can’t end in 3 or 7,
so 5dn is one of {18, 98},
which makes the palindrome **10ac = 88**.

3ac ends in 1 or 9 and the middle digits of 3ac and 3dn must be different, or 4dn and 9ac would be the same; so if 3ac ends in 1 then 3dn = 3ac - 3 and ends in 8, while if 3ac ends in 9 then 3dn = 3ac + 3 and ends in 2. For 3ac, the squares ending in 1 or 9 are {121, 169, 289, 361, 441, 529, 729, 841, 961}, with corresponding digit sums of {4, 16, 19, 10, 9, 16, 18, 13, 16}; eliminating the non-squares from that leaves 3ac in {121, 169, 441, 529, 961}. For 18dn, the numbers that are both happy and lucky are {13, 31, 49, 79}, which makes 18dn + 26 = 9ac one of {39, 57, 75} (not 105, which is too big). That eliminates 121 for 3ac because it would make 3dn 118, and 9ac can’t start with 1, so 3ac is reduced to {169, 441, 529, 961}.

For 8dn, the two-digit square-pyramidal numbers are {14, 30, 55, 91}. We can eliminate 91 because 8dn < 4dn, whose first digit is one of {2, 4, 6} (from 3ac).

Now 11ac starts with 2 or 8 (from 11dn above),
ends with 2 or 8 (from 3dn), has 5 as the second digit,
and its third digit is one of {0, 4, 5} (from 8dn).
None of the possibilities can make its digit sum 10,
so **21dn = 20** and
11ac is one of {2558, 8552},
with **8dn = 55**.
For 4dn to be > 8dn, it must start with 6,
making 3ac one of {169, 961}.
That narrows 3dn to {172, 958}
and 9ac to {75, 57} respectively.

For 1ac, the triangular numbers with 5 in the middle are {153, 253, 351}. The digit product of 253 is 30, not a triangular number, so 1ac is one of {153, 351}. 7ac has a digit sum of 10 and its last digit is even (from 2dn), so it’s one of {28, 46, 64, 82}. Of the 8 possibilities for 1dn, the only ones where the digit sum plus digit product is a multiple of 5 are {12, 38}, narrowing 7ac to {28, 82} and 2dn to {525, 585}.

If 1dn is 12, the instruction from last digits starts with one of the letters [BLV]; otherwise it starts with [HR]. Either 3ac = 961, 3dn = 958, 9ac = 57, 4dn = 67, 5dn = 18, 11ac = 2558 and 11dn = 25; or 3ac = 169, 3dn = 172, 9ac = 75, 4dn = 65, 5dn = 98, 11ac = 8552 and 11dn = 81. In the first option, the last digits of the first seven down answers are (2 or 8, 5, 8, 7, 8, ?, 5), giving letter choices of [BLVHR][EOY][BLV][GQ][BLV]_[EOY] for the first word of the instruction; but there are no seven-letter words with one of the combinations [BLV][GQ][BLV] in the middle (few of which are even pronounceable). The second option has the digits (2 or 8, 5, 2, 5, 8, ?, 5) and letter choices [BLVHR][EOY][BLV][EOY][HR]_[EOY]; possible words are {loverly, reverie, reverse, reverso} and the only one likely to start an instruction is REVERSE.

On that basis,
we must have **1dn = 38**
(making **1ac = 351** and **7ac = 82**),
**2dn = 525**,
**3dn = 172**
(making **3ac = 169**, **9ac = 75**,
**11ac = 8552**, **11dn = 81**
and **14ac = 135**),
**4dn = 65**, **5dn = 98**.
18dn is 26 less than 9ac,
so **18dn = 49**.
6dn is _89 and the prime factors of the possible values are
{189 = 3×3×3×7,
289 = 17×17,
389 = 389,
489 = 3×163,
589 = 19×31,
689 = 13×53,
789 = 3×263,
889 = 7×127,
989 = 23×43}.
Clearly, the prime 389 only has two factors, 1 and itself;
the ones with two primes p and q have only four factors,
{1, p, q, p×q};
that leaves **6dn = 189** with the eight factors
{1, 3, 7, 3×3, 3×7, 3×3×3,
3×3×7, 3×3×3×7}.

The digit product of 22dn is a power of 2,
so 22dn can only contain digits from {1, 2, 4, 8}.
27dn is a number in the Fibonacci series,
ie one of {13, 21, 34, 55, 89}.
Thus 26ac starts with one of [1248] and ends with [12358];
its digit product is a cube, which can only be one of {1, 8, 64},
so 26ac has to be one of {11, 18, 42, 81, 88}.
But 11 is 23ac, 81 is 11dn and 88 is 10ac,
which narrows 26ac to {42, 18}.
If it’s 42 then 27dn is 21,
but then the maximum possible digit sum for 29ac
is 9 + 9 + 1 = 19, not 20 (21dn).
Therefore **26ac = 18** and **27dn = 89**.
Now 29ac is __9 with the middle digit being [1248];
to make a digit sum of 20, the first two digits must add up to 11,
so 29ac is one of {389, 749, 929}.
Only one of those has a digit product that’s a cube,
so **29ac = 389**.

13ac is _9, so its first digit must be even
and it’s one of {29, 69}
(not 49, which is 18dn, or 89, which is 27dn).
13dn is then one of {21 = 7×3, 28 = 7×4, 63 = 7×9}.
20ac is _2_9 (with a power of 2 for the third digit),
so its digit sum is in the range [13, 28]
(1 + 2 + 1 + 9 and 9 + 2 + 8 + 9)
and must be a multiple of 7,
because numbers that are ≥ 13 can’t be a factor of 13dn/7,
which is ≤ 9;
so 20ac’s digit sum is one of {14, 28}
(not 21, because then 20ac would be a multiple of 3, not a prime).
The only numbers giving those digit sums are {1229, 2219, 9289};
the last two are both multiples of 7, not primes,
so **20ac = 1229**, making **22dn = 218**.
As a multiple of 20ac’s digit sum of 14, 13dn must be even,
so **13dn = 28** and **13ac = 29**
(with a digit product of 18, which is 26ac).

17ac is now _84, so its digit sum is in the range [13, 21];
the only triangular numbers in that range are {15, 21},
so 17ac is one of {384, 984} respectively.
If 12dn ends in 9 and has a single-digit digit product,
it must be 19, but the digit product is 9, not even;
so **17ac = 384**.
12dn is _3 and needs an even first digit to make the digit product even,
so **12dn = 23**
(not 43 or higher because then the digit product would be ≥ 12).

To have a digit product of 20,
19ac must be one of {45, 54}.
That makes 15dn one of {34, 35},
with a digit product of 12 or 15,
so we need 12 or 15 as another entry in the grid.
The only places where it could fit are
25ac, 24dn (where we have 1_) and 25dn.
We can rule out 25ac because neither 12 nor 15 is prime;
and neither of the digit products 2 and 5 is square,
so we can’t put 12 or 15 at 24dn;
therefore 25dn is one of {12, 15};
but 12 isn’t a lucky number,
so **25dn = 15**, **15dn = 35**
and **19ac = 45**.

Now 19dn is 41_ and its digit product must be ≥ 10, to be another grid entry, so it’s in the range [413, 419] with possible digit products of {12, 16, 20, 24, 28, 32, 36}. As above, 12 and 16 can’t fit anywhere (25ac is prime; 24dn needs a square digit product); there’s also no room for 24, 32 or 36, so 19dn is one of {415, 417} with its digit product being either 20 (21dn) or 28 (13dn). 28ac is _5_ with a digit product of 180 = 2×2×3×3×5, which can only be achieved with the digits {4, 5, 9} or {5, 6, 6} in some order. 24dn has a square digit product, so it can only be one of {14, 19}, making 28ac one of {459, 954}.

For the last digits of the last seven down entries (19dn to 27dn) we now have
([57], 4, 0, 8, 9, 5, 9) if 28ac is 459,
or ([57], 4, 0, 8, 9, 5, 9) if 28ac is 954.
The corresponding letter options are [EOYGQ][IS][JT][HR][DNX][EOY][IS]
or [EOYGQ][DNX][JT][HR][IS][EOY][IS];
for the first, the only pronounceable sequence would be
from [EOYGQ][IS]THN[EOY][IS], but there are no matching words.
For the second, we can safely discount anything starting with [GQ][DNX]
or containing [DNX]J[HR],
which leaves [EOY][DNX]T[HR][IS][EOY][IS]
and the only available word to end the instruction is ENTRIES.
That makes **19dn = 415** and **24dn = 19**,
with **28ac = 954** and 20dn ending in 4.

16dn is a prime matching 5_, ie one of {53, 59}.
For the middle word of the instruction,
the last digits of entries 11dn to 18dn are (1, 3, 8, 5, [39], 9),
giving letter choices of [AKU][CMW][HR][EOY][CMWIS][IS],
and the only matching word is ACROSS,
making the instruction “reverse across entries”
and **16dn = 59**.

The only remaining gap in the grid is the second digit of the prime 25ac,
which is one of {13, 17}
(as 11 and 19 are already in the grid elsewhere).
If 20dn is 134, its DS + DP is 8 + 12 = 20;
for 174 it’s 12 + 28 = 40;
either way, the result is a multiple of 20 as required.
We know that reversing the across entries must give
a second solution that satisfies the clues,
and reversing 25ac will give either 35 or 75 for 25dn;
but 35 isn’t a lucky number, so we need to end up with 75,
meaning that **25ac = 17** in the initial solution
and the grid is complete, as shown here.

Finally, all the across entries must be reversed. That won’t change any of the properties based on their digit sums or products (as they still have the same digits), but we can check that the other properties, and those of the altered down entries, still hold.