Solution
= Solution
Take $g\in G$. Since $K$ is a <normal subgroup>, $gPg^{-1}\leq K$; it has the same order as $P$, so it is also a <Sylow subgroup> of $K$. The conjugacy assertion in the <Sylow theorems> supplies $k\in K$ such that
$$
k(gPg^{-1})k^{-1}=P.
$$
Consequently $kg\in N_G(P)$, and therefore
$$
g=k^{-1}(kg)\in K N_G(P)=N_G(P)K.
$$
Thus $G\subseteq N_G(P)K$, while the reverse inclusion is immediate because both factors are subgroups of $G$. Hence
$$
\boxed{G=N_G(P)K}.
$$