HK us the group composed of the products of the elements in H and K under multiplication in G, but it is not the direct product (that would be so much easier).
I have figured out the normal subgroup part already, but I am stuck when it comes to nilpotence. I know that H and K are both subgroups of HK but I don't know if that is helpful here.
Set $O_p(G)=\bigcap P$ where $P\in Syl_p(G)$. Notice that $O_p(G)$ is the largest normal $p$ group in $G$.
Now set $$F(G)=\prod O_p(G) $$
It is east to see that $F(G)$ is nilpotent as its Sylow$p$ subgroups are normal.
Now, Let $N,M$ be two normal nilpotent groupx in $G$ and $Q\in Syl_p(N)$. $Q$ in Char in $N$ and $N$ is normal in $G\implies Q$ is normal in $G$. Hence $Q\leq O_P(G)$ .
Hence, we have $N\leq F(G)$ with similiar argument $M\leq F(G)$. Thus, $MN\leq F(G)$. Since $F(G)$ is nilpotent then $MN$ is nilpotent.