 Solution

Set $A=\alpha a^{†}$ and $B=-{\alpha }^{\ast }a$. Their commutator isIt commutes with both operators, so the stated Baker--Campbell--Hausdorff formula terminates:In particular $e^{-{\alpha }^{\ast }a}|0⟩=|0⟩$. Expanding the remaining exponential and using $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$ gives the number state expansionThe orthonormality of the number states givesThis is a Poisson distribution with parameter $\mu =|\alpha {|}^2$. Its probabilities sum to $e^{-\mu }{e}^{\mu }=1$. The phase of $\alpha$ affects the coherent state's relative probability amplitudes but not its photon number probabilities. At $\alpha =0$ the distribution is concentrated at $n=0$.