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sorry guys.... you got it all wrong...
answer is D.
A is sufficient as everybody noticed....
but B is sufficient as well... here goes:
the important part is that p is odd (!!!) as given in the stem. so if it is a sum of two squares one must be odd and the other must be even....
but - an even square is always divisible by 4 ( since (2k)^2=4k^2), and an odd square always leaves a remainder of 1 when divided by 4 since (2k+1)^2=4k^2+4k+1
hence their sum (which is p) has a remainder 1 when divided by 4.