Find an angle in the figure defined by a equilateral triangle and a regular pentagon

Note that $BM=BE=BD$ so $B$ is the circumcenter of $\triangle DEM$. It follows that $$\angle EDM = \frac 12 \angle EBM = \frac 12 \cdot 60^\circ = 30^\circ.$$ Now, $$\angle MDC = \angle EDC - \angle EDM = 108^\circ - 30^\circ = 78^\circ.$$ By symmetry $\angle DCM = 78^\circ$. Therefore $$\angle CMD = 180^\circ - \angle MDC - \angle DCM = 24^\circ.$$

Hint: without loss of generality, we can set $MN=1$, then the altitude $MK$ of $\triangle{NMP}=\frac{\sqrt{3}}{2}$. This altitude will also be the altitude of the $\triangle{CMD}$ and the bisector of the angle in question. Now using law of sines and $\triangle{NBC}$, we can obtain that $$CD=\frac{1}{1+4\frac{sin \: 48°}{\sqrt{3}}}$$ Now that we know base and altitude in the $\triangle{CMD}$, it's not difficult to find the required angle.