Let $P$ be a sylow p-subgroup of $G$ and let M be any subgroup of $G$ which contains $N_{G}(P)$. Then how should we approach to show $|G:M|\equiv 1$ (mod p)?
My attempt : From Sylow's Theorem we can easily observe that the number of Sylow $p$-subgroups of a group $G$ is equal to $|G:N_{G}(P)| \equiv 1(\text{mod } p)$ and here $M$ contains $N_G(P)$. So, $P$ is also a Sylow $p$ subgroup of $M$.
After this point I am not able to understand how to proceed further. Any help/hint in this regards would be highly appreciated. Thanks in advance!
Note that $|G:N_G(P)|=a_G$ is the number of Sylow $p$-subgroups of $G$. As $M\supseteq N_G(P)$, then $N_M(P)=N_G(P)$ and so $|M:N_M(P)|=|M:N_G(P)|=a_M$. But $a_G\equiv a_M\equiv1\pmod p$, etc.