 Solution

A Sylow subgroup for the prime $p$ is a maximal $p$-subgroup; in a finite group of order $p^{a}m$ with $p\nmid m$ it has order $p^a$. Sylow theorems assert existence, containment of every $p$-subgroup in a Sylow $p$-subgroup, conjugacy of all Sylow $p$-subgroups, and the counting conditions $n_p\mid m$, $n_p\equiv 1(\mathrm{mod}p)$.

Since $N$ is a normal subgroup, $NP$ is a subgroup andThe final index is prime to $p$. Since $N\cap P$ is a $p$-subgroup, its index in $N$ being prime to $p$ means it contains the entire $p$-part of $|N|$. Hence $N\cap P$ is Sylow in $N$.