The intersection of a normal subgroup and Sylow $p$-subgroup

You're almost there.

So, you have chosen a $P_0\in Syl_p(H)$ such that $P\cap H\le P_0$. Then, we have $P_0\le gPg^{-1}$. Also, $P_0\le H$, so, as $gHg^{-1}=H$, it means $$g^{-1}P_0\,g\le P\cap H\,.$$ Assuming everything is finite, by calculating sizes, we are ready, as $|g^{-1}P_0\,g|=|P_0|$ and both are included in $H$, so $|P\cap H|=|P_0|$ also follows.

Hint: $PH$ is a subgroup when $H$ is a normal subgroup. Apply the formula