If $K\unlhd G$, $G$ is finite, $P\in{\rm Syl}_p(K)$. Show that $G=N_G(P)K$ Let $|G|=p^{\alpha}.m$ where $p$ is prime. Since $K\unlhd G$, we have $P\unlhd G$.Hence $P$ is the unique Sylow $p$ subgroup of $G$. Thus $K=P$. $N_G(P)$ is the stablizer, so by orbit stabilizer theorem, $|G|=|P||N_G(P)|$.
Is my argument correct?
 A: The following is true and it is actually more than you are asking for. But seldomly one finds this in a textbook. All groups are assumed to be finite.

Theorem Let $N \unlhd G$, and $P \in Syl_p(N)$. Then 
(a) ($\color{blue}{Frattini \ Argument})\ G=N_G(P)N$.
(b) $|G:N_G(P)| \equiv 1$ mod $p$.
(c) There is an $S \in Syl_p(G)$ with $P=N \cap S$ and $S \subseteq N_G(P)$.
Proof (a) Let $g \in G$. Since $N$ is normal, the conjugate $P^g \subseteq N^g=N$. Of course $|P|=|P^g|$ and hence $P^g \in Syl_p(N)$. So by Sylow Theory in $N$, we can find an $n \in N$ such that $P^n=P^g$. This is equivalent to $gn^{-1} \in N_G(P)$, and thus $g \in N_G(P)N$.
(b) From (a) we have $G/N \cong N_G(P)/(N \cap N_G(P))$. Note hat $N \cap N_G(P)=N_N(P)$, and $|N:N_N(P)|$ is the number of Sylow $p$-subgroups of $N$. Hence $|N:N_N(P)|=|G:N_G(P)| \equiv 1$ mod $p$.

(c) Since $P$ is a $p$-subgroup of $N_G(P)$, by Sylow Theory in $N_G(P)$, we can find an $S \in Syl_p(N_G(P))$, with $P \subseteq S \subseteq N_G(P)$. We will argue that in fact $S \in Syl_p(G)$. Note that $P \subseteq S \cap N$. On the one hand $|S \cap N :P| \mid |S:P|$, which is a $p$-power. But on the other hand, $|S \cap N:P| \mid |N:P|$, which is a $p'$-number since $P$ is Sylow in $N$. So we must have $P=S \cap N$. By (b) we have $|G:N_G(P)|$ is not divisible by $p$. Since $S$ is Sylow in $N_G(P)$, it follows that $S$ is Sylow in $G$. 
