Verified answer

First of all, H∩K is the intersection of two subgroups. The normality of H is irrelevant here. H∩K is a subgroup by the "quick" subgroup test:

If a,b are in H∩K, then ab⁻¹ is in H (since H is a subgroup) and K (since K is a subgroup). Thus H∩K is a subgroup. Easy enough!

It is also a subgroup of K -- it is contained in K (obviously, that's just the definition of an intersection of sets) and it is a subgroup. Why is it normal? Well, consider a cosets:

(H∩K)a

a(H∩K)

Note: a is in K, since we are considering these as cosets in K.

These cosets are made up of elements of H right? And because of that, we know that when a acts on these (with respect to H) they do not change left-or-right because H is normal. And when a acts on these (with respect to K) they don't move around because a is in K.

More formally, you just rely on the fact/theorem that:

a(H∩K) = (aH)∩(aK)

and finish by using the facts that are known:

aH = Ha (H is normal)

aK = K = Ka (since a is in K)

Thus:

a(H∩K) = (aH)∩(aK) = (Ha)∩(Ka) = (H∩K)a