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Let $z_0\in\Bbb{C}$ be arbitrary. The matrices $A_0:=A(z_0)$ and $B_0:=B(z_0)$ are normal; in view of $A_0^*=A^\#(z_0)$ and $B_0^*=B^\#(z_0)$ they commute with their Hermitian adjoints. They also commute with each other. So $A_0$ and $B_0$ could be simultaneously diagonalized by a unitary matrix. That matrix also diagonalizes $A_0^*$ and $B_0^*$. So the four matrices $A_0=A(z_0),B_0=B(z_0),A_0^*=A^\#(z_0),B_0^*=B^\#(z_0)$ could be simultaneously diagonalized which implies that they all commute with each other. So $B(AA^\#)=(AA^\#)B$ holds throughout $\Bbb{C}$.