Here are two different proofs. $K \in Syl_p(G)$ and put $H=N_G(N_G(K))$.
$(1)$ Since $N_G(K) \unlhd H$ and $K \in Syl_p(G)$, whence $K \in Syl_p(N_G(K))$, we can apply the Frattini Argument, which gives $H=N_H(K)N_G(K)=(H \cap N_G(K))N_G(K)=N_G(K)N_G(K)=N_G(K).$
The second proof relies on a counting argument.
$(2)$ Note that $K \ char \ N_G(K) \unlhd H$, it follows that $K \unlhd H$. So $H$ has a unique Sylow $p$-subgroup. Hence $n_p(H)=|H:N_H(K)|=|N_G(N_G(K)):N_G(K)|=1$ and again the result follows.
And here is a bonus:
Proposition Let $G$ be a finite group, $P \in Syl_p(G)$, then every subgroup between $N_G(P)$ and $G$ is self-normalizing, that is, if $H$ is a subgroup of $G$ with $N_G(P) \subseteq H \subseteq G$, then $N_G(H)=H$.