How $H \cap P$ is a Sylow $p$-subgroup of $H$?

I searched the results relating to this and I found the following one:

Let G be a finite group, $H \triangleleft G$ be a normal subgroup and $P$ be a Sylow $p$-subgroup of $G$. Then $H \cap P$ is a Sylow $p$-subgroup of $H$ and its proof is found here.

So the answer for my question " What is the importance of this result if any " is $$\text{when $H$ is normal, then the statement in the OP is true!} $$ That is, normality is essential in the hypothesis of H

Even a more general fact holds true:

If $G$ is a finite group, $H \lhd \lhd G$ (subnormal) and $P \in Syl_p(G)$, then $H \cap P \in Syl_p(H)$.

Proof Since $H \lhd \lhd G$, we can find a subnormal series: $H=H_0 \lhd H_1 \cdots \lhd H_i \lhd \cdots \lhd G=H_r$. Since $H_{r-1} \lhd G$, $P \cap H_{r-1} \in Syl_p(H_{r-1})$. But $H_{r-2} \lhd H_{r-1}$, so $P \cap H_{r-1} \cap H_{r-2}=P \cap H_{r-2} \in Syl_p(H_{r-2})$. Now induction finishes the proof.