If N is even/odd, we need an even/odd number of odd numbers in the sum. This already halfes the number of possible solutions. E.g. for N=14, there is no point in checking any combinations where P is odd.
Rewriting the formula given, we get:
N = A + (A+1) + (A+2) + ... + (A+P-1)
= P*A + 1 + 2 + ... + (P-1)
= P*A + (P-1)P/2 *
= P*(A + (P-1)/2)
= P/2*(2*A + P-1)
The last line means that N must be divisible by P/2, this also rules out a number of possibilities. E.g. 14 only has these divisors: 1, 2, 7, 14. So possible values for P would be 2, 4, 14 and 28. 14 and 28 are ruled our for obvious reasons (in fact, any P above N/2 can be ignored).
This should be a lot faster than the brute-force approach.
(* The sum of the first n natural numbers is n(n+1)/2)