Let $H$ be a nilpotent group and $K$ a supersolvable group. Considering $G=H \times K$, then $G$ is supersolvable.
I tried to create a normal cyclic series, but I don't know how to proceed, maybe the idea is to create a specific one. If group H were finite, it would be supersolvable and the exercise would follow.
Even abelian groups need not be supersolvable. For instance, the additive group of rational numbers $\Bbb Q$ is not supersolvable. So nilpotent groups need not be supersolvable. In particular $G=H\times K=\Bbb Q\times 1$ is not supersolvable, although $K=1$ is supersolvable and $H=\Bbb Q$ is nilpotent.
I suspect that you mean the claim of the duplicate below (where the product is $G=HK$ of finite groups, and with normal subgroups).
Prove that product of a nilpotent group and a supersolvable group is a supersolvable group