This is one possible solution path. As they are the reverse of primes, all entries start with one of {1, 3, 7, 9} and A, B, D, F, H and M consist of those digits only. The two-digit entries in that list are B, F and M, for which the possible values are {13, 17, 31, 37, 71, 73, 79, 91, 97} (not 11, because it would have the same digit in adjacent cells); note that these are all prime numbers except for 91 = 7×13.

From O = NUN,
N^{2} has no more than 3 digits,
so N < 32.
From I = (B + B)/N + Y, N is a factor of 2B.
If B is prime then the only factors of 2B are {1, 2, B, 2B};
N can’t be 1, 2 or B, so N would have to be 2B,
which means B < 16, ie, B = 13 and N = 26.
But 26 isn’t the reverse of a prime,
so B isn’t prime.
Therefore **B = 91**.
That gives 9 for the middle digit of D
and restricts its first digit to 3 or 7 (having neighbours of 9 and 1).

The factors of 2B = 2×7×13 are {1, 2, 7, 13, 14, 26, 91, 182},
of which N can only be 13 or 14.
If N = 13 then F = 17 (the only other available value starting with 1),
which would make I = 73,
and then Y = I − 2B/N = 73 − 2×91/13 = 59.
D = YT is a multiple of Y,
but the only multiples of 59 in the 300s or 700s are {354, 708, 767},
none of which matches D, so **N = 14**.

F is 13 or 17.
If F is 13 then (from the grid) I is 34,
which makes Y = 34 − 2×91/14 = 21.
The only odd multiples of 21 in the 300s or 700s are {315, 357, 399, 735, 777},
none of which matches D (399 does, but it has two adjacent digits the same),
so **F = 17**, **I = 74** and **Y = 61**.
The only odd multiples of 61 in the 300s or 700s are {305, 793},
so the only match is **D = 793**, giving **T = 13**.

A = D + O/W tells us A > D,
and as it has the same first digit and contains only digits {1, 3, 7, 9},
it must be **A = 797**, with W = O/4.

H is now 71_ and its last digit is one of {3, 7, 9},
which also starts O.
We know O = 196U.
The only multiples of 196 in the 300s, 700s or 900s are {392, 784, 980};
980 isn’t allowed (it ends in zero);
784 would make H = 717,
in which the sum of digits is a multiple of 3,
so 717 (forwards or backwards) is a multiple of 3, not a prime.
Therefore **O = 392**, **U = 2**, **H = 713** and **W = 98**.

From H = SC/F we now have SC = FH = 12121,
so S and C are both odd, not ending in 5,
ie each of them ends in one of {1, 3, 7, 9}.
Thus C is in the range [313, 397]
(its middle digit can’t be 0 because it’s the last digit of G),
so S is between 12121/397 ≈ 30.5 and 12121/313 ≈ 38.7,
ie it’s one of {31, 33, 37}.
The only one that gives an integer result is **S = 31**, with **C = 391**.
In the grid,
M is 7_ but its second digit has neighbours of 1, 7 and 9,
so **M = 73**.

For M = J + J − G + L,
we have M = 73, G ending in 9 and L ending in 2,
so 2J = M + G − L ends in 0.
As J itself can’t end in 0, it must end in 5,
so **J = 35**, completing **L = 152** in the grid.
Then G = 2J + L − M, so **G = 149**.

E = (K − L + U − T)Z now reduces to E = (K − 163)Z.
K is 3_9 in the grid;
its middle digit can’t be any of {0, 3, 6, 9}
because they’d give a digit sum that’s a multiple of 3,
so the entry wouldn’t be the reverse of a prime,
and it can’t be any of {2, 3, 4, 5, 7, 9} because those digits occur in neighbouring cells;
so K is 319 or 389 and
E (9_4 in the grid) is a multiple of 156 or 226, respectively.
The only multiple of 156 in the 900s is 936, not a match for E,
so **K = 389**.
The only multiple of 226 in the 900s is 904,
so **E = 904**, **Z = 4** and the puzzle is complete.