 Solution

The cyclohexanone enamine attacks the beta carbon of the cyclopentenone. Its nitrogen lone pair forms an iminium C=N bond as the enamine pi bond forms the new C–C bond; simultaneously the enone pi system shifts onto its oxygen. proton transfer converts the resulting cyclopentanone enolate back to the ketone.

A final iminium deprotonation regenerates an enamine, but the position of its double bond matters. Label the nitrogen-bearing cyclohexane carbon $h_1$, its initially nucleophilic neighbour $h_2$, and the other neighbour $h_6$. After addition, $h_2$ bears the cyclopentanone substituent. deprotonation at $h_6$ gives $h_1=h_6$, the less substituted, less hindered enamine drawn in A. Merely restoring the original $h_1=h_2$ bond would give a different regioisomer. Interconversion through the iminium ion permits this enamine regioisomerization under the reaction conditions. The methyl and mesylate substituents are spectators during this first Michael addition.