Let $I$ be the inventer of $\Delta ABC$. It is known that $I$ is also the orthocenter of $\Delta M_A'M_B'M_C'$, so $AM_A'\perp M_B'M_C'$, and so $AP_A$ is the angle bisector of $\angle BAC$. Therefore, by angle bisector theorem $$\prod_{cyc} \frac{M_BP_A}{P_AM_C}=\prod_{cyc} \frac{M_BA}{AM_C}=\prod_{cyc}\frac{CA/2}{AB/2}=1$$Therefore, by ceva's theorem on $\Delta M_AM_BM_C$ $M_AP_A$, $M_BP_B$ and $M_CP_C$ concur

From I-E Lemma if you let $I$ the incenter of $\triangle ABC$ then $M_B'M_C'$ is perpendicular bisector of $AI$ which means $AP_A$ bidects $\angle BAC$ and now from homothety on the ratios and the fact that $AM_CM_AM_B$ is a parallelogram we can easly conclude the point of concurrency will in fact be the Isotomic Conjugate of the incenter of $\triangle M_AM_BM_C$, thus done .