It is the equation 2.5 in the theorem 1 of the paper Hall.
I am mentioning the theorem below-
Theorem ([M. Hall]) Let $K \unlhd G$, $P \in Syl_p(G)$, then $n_p=a_pb_pc_p$, where
$a_p = \#Syl_p(G/K)$
$b_p=\#Syl_p(K)$ and
$c_p=\#Syl_p(N_{PK}(P\cap K)/(P \cap K))$.
If you can't see the paper, just consider just this equality in proof of the theorem, as this is the only part i don't get.
$\textbf{Problem-}$ If $G$ is a finite group and $K\triangleleft G$ and $P$ is a sylow $p-$ subgroup of $G$, then he writes this equation,
$P\cup (N_{PK}(P)\cap K)=N_{PK}(P)\cap (P\cup K)=N_{PK}(P).$
Now out of these 2 equalities, first one is clear to me, but problem is, how he is writing the second equality i.e. $N_{PK}(P)\cap (P\cup K)=N_{PK}(P)$.
I have seen it from alot many many angles, discussed it with my professor, cant write all my attempts here, have spent many hours on it, but nothing. please help. Any help is appreciated.
So, what he is saying is this: $P(N_{PK}(P) \cap K)= N_{PK}(P) \cap PK= N_{PK}(P)$, use modular law* for first equality. Since $P \subseteq N_{PK}(P)$, $N_{PK}(P) \cap K \triangleleft N_{PK}(P)$ and $P\cap (N_{PK}(P) \cap K)= P\cap K $, this gives $N_{PK}(P)/N_{PK}(P) \cap K \cong P/P \cap K$.
Modular law: if $pk=x$ with $p \in P$, $k \in K$ and $x \in N_{PK}(P)$, then $k = p^{-1}x \in N_{PK}(P)$ so $k \in N_{PK}(P) \cap K$.