IMO C

i just tried by plugging sme selected no's( and a bit of assumptions)

to meet s1) we need p+n = 5k+1(where k is int)
there can be n no. of combination hence
s1) alone insuff

to meet s2) we need p-n=3K+1

same as s1) insuff...

combining
u will come to knw that no,s like p=5 and n=1; p=20 and n=1are satisfying bth the statements
and remn is 9 for bth the cases

now time to assume that it holds true for all such no's

hence C

sorry for such a vague reply
but on exam such tricks will play a role of crocin to keep your headache away

in a nutshell,
once u realize neither a nor b is a right answer,
try to expand the equasion as ((p - n)(p+n))/3*5 >>>>>>>>>>>>> then plug (6+5)/5 r 1 and (6-5)/3 r 1, then, plug (4+2)/5 r 1 and (4-2) r 2.
Since there is no consistency - E

Please let me know if there is something faulty in this approach. Thanks