$H ∩ N$ is not necessarily a normal subgroup of $N.$
Take $S_3$ for example
$S_3$ is the normal subgroup of $S_3.$
and $(e,(1,2))$ is the subgroup of $S_3.$
So, let $N = S_3$
and $H = (e, (1,2))$
Then we find that $H∩N = H = (e,(1,2))$ which is not a normal subgroup of $N.$ (Think)
So, $H∩N$ need not necessarily be a normal subgroup of $N.$
So, either option B or option D is true.
Now, consider the theorem
$N$ is a normal subgroup of $G$ iff $gNg^{-1} ⊆ N$ for all $g∈G.$
Now, let $x$ be an arbitrary element in $N.$
then $gxg^{-1} \in N.$
Now, $H$ is a subgroup of $G.$
So, for all $g∈ H$ and $x$ an arbitrary element in $H,$
we have $gx∈ H.$
Again from the definition of a subgroup we have
$gxg^{-1} ∈ H.$
So, we get that for any arbitrary element $x$ in $H∩N$
$(gxg^{-1}) ∈ H∩N$ for all $g∈ H.$
Hence $(g H∩N g^{-1} ) ⊆ H∩N,$ for all $g ∈ H$
So, $H∩N$ is a normal subgroup of H.
Thus, Option B is the correct answer.