= Solution
A <Sylow subgroup> for the <prime> $p$ is a maximal $p$-<subgroup>; in a <finite group> of order $p^am$ 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 $n_p\mid m$, $n_p\equiv1\pmod p$.
Since $N$ is a <normal subgroup>, $NP$ is a <subgroup> and
$$
[N:N\cap P]=[NP:P]\mid[G:P].
$$
The 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 \b[$N\cap P$ is Sylow in $N$].
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