Pentomino Primes was solved by Joseph W. DeVincentis and Lee Kaplan ("This was a great puzzle.")

The solution to the puzzle is:

23452
43214
15521
31433
12455
45214
13351
32422
25315
15433
41354
52142

A=2, B=5, C=11, D=13, E=17, F=23, G=29, H=31
J=37, K=41, L=43, M=47, N=53, P=61, Q=79, R=83

Notes from the original puzzle by John Gowland.  I changed the letters of the puzzle to make it a bit easier.

1.  Working backwards, XII = 2DN2 = 2.31.292 = 52142. Thus D=31 and N = 29
2.  XI =2DNQ = 2.31.29.23.= 41354. Thus Q = 23
3.  X = PQC = .(3,4,5). Only solution with 3,4,5 in the middle is 61.23.11. Thus P = 61 and C = 11
4.  IX = FPK = F.61.K. It must start with a 2 or 3 and not end in 3 or 4.
     67.61.13 = 53131 and 83.61.7 = 35441. It cannot end in ...41 as there is already a 4 in the
     Y-pentomino. Only solution is 83.61.5 = 25315 giving F = 83 and K = 5.
5.  VIII = 2GNB = 2.G.29.B. It must start with a 3 and end with a 2. Only satisfactory solution is
     2.43.29.13 = 32422. Thus G = 43 and B = 13
6. VII ends with 1 or 5. VII = MB2 = M.132 = 79.132 = 13351. Thus M = 79
7. VI begins with a 4 and cannot contain 3 in the last 3 digits. 2JLB = 2JL.13 ; of the five possible
     solutions, only 2.47.37.13 = 45214 begins with a 4.  Thus J = 47 and L = 37
8.  V = HJK and ends with a 5 = H.47.5 = 53.47.5 or 61.47.5. But P =61, so H =53 and HJK = 12455
9.  IV = G2E = 432.E and begins with a 3. E could be 17 or 19. However, 432.19 = 35131 and the
     T-pentomino already contains a 5. Therefore, E = 17 and 432.17 = 31433
10. III = FEC = ...2. = 83.17.11 = 15521
11. II = 2ADE = 2.A.31.17 making A = 43 or 41.
      But G =43, therefore A = 41 and 2ADE = 41.13.17 = 43214
12. Check I = 4ABC = 4.41.13.11 = 23452.