$G$ is a finite group, $H$ normal in $G$ and has order $p$ prime. Let $K$ be a subgroup of $G$. If $\gcd (| K |, p-1) = 1$ then $HK$ is abelian. 
Let $G$ be a finite group, $H$ is a normal subgroup in $G$ and has order $p$ prime. Let $K$ be a subgroup of $G$. If $\gcd (| K |, p-1) = 1$ then $HK$ is abelian.

I believe that because of $\gcd (| K |, p-1) = 1$, I would have to come up with some application of the Lagrange theorem.
I do not know how to relate something abelian with the Lagrange theorem, I also tried the order of $HK$, which is $| K | \cdot p$, and with that arrive at some equation of the type
$(p-1) x + y \cdot | K | = 1$ to use the fact that $gcd (| K |, p-1) = 1$, but I did not get anything.

In general, your statement is not true, see the example in the comment of carmichael561. If $H \lhd G$ and $|H|=p$, then a subgroup $K$ acts by conjugation on $H$ as automorphisms. The kernel of this action is $C_K(H)=\{k \in K: hk=kh, \text{ for all } h \in H \}$. Hence there is a homomorphic embedding $K/C_K(H) \rightarrow Aut(H) \cong C_{p-1}$. But $gcd(|K|, p-1)=1$, implying $K=C_K(H)$, equivalently, $K \subseteq C_G(H)$ (note that this implies that $K \unlhd HK$). Since $H$ is abelian, also $H \subseteq C_G(H)$. So the only conclusion you can draw is that $HK$ centralizes $H$. Not that $HK$ is abelian.
You can take it a small step further: since $|H|=p$, we must have $H \cap K=1$ or $H \cap K=H$. In the latter case, $H \subseteq K \subseteq C_G(H)$. In the first case, since $K$ is normal in $HK$, we must have $HK \cong H \times K$.
