By hypothesis, [G: N] = 50. So N is not equal to G. Let H be a subgroup such that N < H < G.

Then P < N < H < G. So N_H(P) < N_G(P) = N. Note that N_H(P) = N:

Pick g in N. Then gPg^-1 = P. Since N < H, it follows that g is in H. So g is in N_H(P) and hence N_H(P) = N.

Now [G: H][H: N] = 50 and since N_H(P) = N, it follows that [H: N] is the number of Sylow 7-subgroups in H. So [H: N] = 1 mod 7 and divides 50.

The divisors of 50 are 1, 2, 5, 10, 25, and 50. Of these, only 1 and 50 are congruent to 1 mod 7.

So [H: N] = 1 or 50. If [H: N] = 1, then H = N. If [H: N] = 50, then [G: H][H: N] = 50 implies [G: H] = 1. So H = G.