Let $G$ be a group with $K\unlhd G$ and Abelian $N\unlhd G$. Is $G/(NK)$ an Abelian group?

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Let $G$ be a group such that $N$ is an Abelian normal subgroup of $G$ and $K$ is just a normal subgroup of $G$, is $G/(NK)$ an Abelian group? Also we know that $NK$ is normal in $G$.

I am not really sure how to show it, assuming it's correct which I don't know for sure. Writing down: $g(NK)h(NK)=g(NK)g^{-1}ghg^{-1}g(NK)g^{-1}$, and I know that also $NK$ is normal in $G$ the last equality means $g(NK)g^{-1}=NK$, but I want to show that: $g(NK)h(NK)=h(NK)g(NK)$.

Confused, how do you suggest me to get unstuck? Thanks!