$Q \cap N_G(P)=Q\cap P$ where $P,Q$ are both Sylow $p-$groups We just need to prove that $Q \cap N_G(P) \subset P$. This problem becomes to show $gPg^{-1}=P, g \notin P, g\in Q \implies g \in P$. So how exactly $g$ being a member of another $p$-Sylow group helps him to be a member of $P$?

Apply Sylow Theory in $N_G(P)$: $Q \cap N_G(P)$ is a $p$-group in $N_G(P)$, and hence must be contained in some Sylow $p$-subgroup of $N_G(P)$. Since $P \unlhd N_G(P)$, $P$ is the unique one and so $Q \cap N_G(P) \subseteq P$. (Note that $Q$ does not even have to be a Sylow $p$-group here, just being a $p$-subgroup suffices.)
