$G/Z(G)$ is cyclic then is abelian? my question is if $G/Z(G)$ is order of p then is commutative then is abelian group but if G is abelian then $G=Z(G)$ therefore $G/Z(G)$ is not order of p?Is it contradiction?

What you would have shown (if you had formulated more carefully) is that $G/Z(G)$ can never be of prime order. This seems to remove any interest from the statement, since its hypothesis is impossible to satisfy.
However this is because the statement is just too limited. If you look at the proof, it actually shows something more general:

If $G$ is a group, $H$ is a central subgroup ($[G,H]=\{e\})$, and $G/H$ is cyclic, then $G$ is Abelian.

In this form there can be useful applications.
Proof: Let $H$ be as stated and $g\in G$ such that $gH$ generates the quotient $G/H$. Then $G$ is generated by the fully commuting set of elements $\{g\}\cup H$, and is therefore and Abelian group. $\Box$
