jmc

Take point $N$ inside $△ABC$ such that $\angle CBN=7^{\circ }$ and $\angle BCN=23^{\circ }$. $\angle MCN=106^{\circ }-2\cdot 23^{\circ }=60^{\circ }$. Also, since $△AMC$ and $△BNC$ are congruent (by ASA), $CM=CN$. Hence $△CMN$ is an equilateral triangle, so $\angle CNM=60^{\circ }$. Then $\angle MNB=360^{\circ }-\angle CNM-\angle CNB=360^{\circ }-60^{\circ }-150^{\circ }=150^{\circ }$. We now see that $△MNB$ and $△CNB$ are congruent. Therefore, $CB=MB$, so $\angle CMB=\angle MCB=\boxed{83^{\circ }}$.