Congruence condition on index of subgroup containing Sylow-normalizer

Statement
Suppose $$G$$ is a finite group, $$p$$ is a prime number, $$P$$ is a $$p$$-Sylow subgroup, and $$H$$ is a subgroup of $$G$$ such that $$N_G(P) \le H$$. In other words, $$H$$ contains a $$p$$-fact about::Sylow normalizer. Then, the index of $$H$$ in $$G$$ is congruent to $$1$$ modulo $$p$$.

Facts used

 * 1) uses::Congruence condition on Sylow numbers
 * 2) uses::Sylow satisfies intermediate subgroup condition: Any $$p$$-Sylow subgroup of a group is also a $$p$$-Sylow subgroup in any intermediate subgroup.
 * 3) uses::Index is multiplicative

Proof
Given: A finite group $$G$$, a prime $$p$$, a $$p$$-Sylow subgroup $$P$$, a subgroup $$H$$ of $$G$$ containing $$N_G(P)$$.

To prove: $$[G:H] \equiv 1 \mod p$$.

Proof: