Let us prove that $AK$ is the bisector of $\angle BAM$. Point $P$ is equidistant from the lines $BC$ and $AC$, and also from the lines $BM$ and $BC$, since $\angle ABM = \angle MBC = \angle CBP = 60^\circ$ (fig. 13). Thus $P$ is equidistant from the lines $BM$ and $MC$, which means that $P$ belongs to the bisector of $\angle BMC$. Therefore $K$ is equidistant from the lines $AC$ and $BM$, and also from $BM$ and $AP$, which proves that $AK$ is the bisector of $\angle BAC$.
Denote $\angle BAK = \angle KAC = \alpha$. Then step by step we can calculate the following:

$$\begin{align*} \angle KMC &= \frac{1}{2} \angle BMC = \frac{1}{2} (60^\circ + 2\alpha) = 30^\circ + \alpha. \\ \angle BCA &= 60^\circ - 2\alpha. \\ \angle BCP &= \frac{1}{2} (180^\circ - 60^\circ + 2\alpha) = 60^\circ + \alpha, \\ \angle ACP &= 60^\circ - 2\alpha + 60^\circ + \alpha = 120^\circ - \alpha. \\ \angle MPC &= 180^\circ - (120^\circ - \alpha + 30^\circ + \alpha) = 30^\circ; \\ \angle AKM &= 180^\circ - \angle AMK - \alpha = \\ &= \angle KMC - \alpha = 30^\circ + \alpha - \alpha = 30^\circ = \angle MPC, \text{ what was to be proved.} \end{align*}$$

Fig. 13