Let $G$ be a finite group and $P$ be a Sylow $p$-subgroup of $G$. If $H,K\trianglelefteq G$, then prove that $$|P\cap HK|=\frac{|P\cap H||P\cap K|}{|P\cap H\cap K|}.$$
Attempt: If $H\cap K=\{e\}$, then the above equality follows from that $P\cap HK$ is a Sylow $p$-subgroup of $HK$ since $HK\trianglelefteq G$ and that the Sylow $p$-subgroup of $HK$ must have form $H'K'$. But when $H\cap K\neq\{1\}$, I don't know how to deal with it. Can anyone give some help?
Hint: As $H, K\unlhd G$ then $P\cap HK\in\text{Syl}_p(HK)$ and $P\cap H\in\text{Syl}_p(H)$. As $K\unlhd KH$ and $H\cap K\unlhd H$ then $(P\cap HK)K/K\in\text{Syl}_p(HK/K)$ and $(P\cap H)(H\cap K)/H\cap K\in\text{Syl}_p(H/H\cap K)$. Using this, a possible approach is the following.
By the second isomorphism theorem it holds $HK/K\cong H/H\cap K$ and hence the Sylow $p$-subgroups of these groups have the same cardinal. The result follows since $$|(P\cap HK)K/K|=\frac{|P\cap HK||K|}{|(P\cap HK)\cap K||K|}=\frac{|P\cap HK|}{|P\cap K|}$$ and $$|(P\cap H)(H\cap K)/(H\cap K)|=\frac{|P\cap H||H\cap K|}{|(P\cap H)\cap(H\cap K)||H\cap K|}=\frac{|P\cap H|}{|P\cap H\cap K|}.$$
