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Is there a finite $p$-group $G$ such that :
(a) $G= \langle A,x,y \rangle$, with $G/Z(G)$ has exponent $p$, $A$ is a maximal abelian normal subgroup of $G$, and $G/A$ has order $p^2$ (thus it is elementary abelian of rank $2$);
(b) the class of $G$ is equal to $p$;
(c) the subgroup $\langle B,y \rangle$ has class exactly $p$, where $B$ denotes the center of $\langle A,x \rangle$.
Thanks in advance
I think the same example works that I gave to your question https://math.stackexchange.com/questions/571949.
Let $H = C_p \wr C_p$, and $G = H_1 \times H_2$ with $H_1 \cong H_2 \cong H$. So $|H|=p^{p+1}$, $|G|=p^{2(p+1)}$.
The maximal abelian normal subgroup $A$ is the direct product of the base groups of $H_1$ and $H_2$, and is elementary abelian of order $p^{2p}$.
Now $H$ has class $p$ and hence so does $G$, which is property (b).
The centre of $H$ has order $p$ and $H/Z(H)$ has class $p-1$ and is generated by two elements of order $p$, so $H/Z(H)$ has exponent $p$ and hence the same applies to $G/Z(G)$, which gives property (a).
We take $x \in H_1$ and $y \in H_2$ to be elements outside of $A$. So $B=Z(\langle A,x \rangle)$ includes the base group of $H_2$, and hence $\langle B,y \rangle$ contains $H_2$ and has class $p$.
