Normalizer of normalizer of Sylow p-subgroup

Question

The question was to show that $P\in Sly_p(G)$ then $N_G(N_G(P))=N_G(P)$, where $G$ is a group. I did it using the Sylow C-theorem but was wondering if it was possible to do the problem without appealing to the C-theorem or $P$ being normal in $N_G(P)$. I did it the following way, $x\in N_G(N_G(P))$ then $xPx^{-1}\leq xN_G(P)x^{-1}=N_G(P).$Since $P,xPx^{-1}$ are Sylow p-subgroups of $N_G(P)$ then for some $y\in N_G(P)$, $xPx^{-1}=yPy^{-1}=P$. Thank you in advance