 Solution

Take $g\in G$. Since $K$ is a normal subgroup, $gP{g}^{-1}\le K$; 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 thatConsequently $kg\in N_G(P)$, and thereforeThus $G\subseteq N_G(P)K$, while the reverse inclusion is immediate because both factors are subgroups of $G$. Hence