My solution: From the condition we get $ PM $ is parallel to the isogonal conjugate of $ AM $ WRT $ \angle BAC $ . From $ AB=AC \Longrightarrow $ the internal bisector of $ \angle BAC $ is perpendicular to $ BC \Longrightarrow BC $ is the external bisector of $ \angle AMP $ , so $ BC $ and the internal bisector of $ \angle BAM, \angle BPM $ are concurrent at $ A- $ excenter of $ \triangle APM $ . Q.E.D

let AT is thr line such that AT and MN are parallel.then use the lemma: in triangle ABC,if AD is internaly bisector of angle BAC then we have :BD/CD=AB/AC

One may also consider the similar triangles $AMB$ and $NMC$ (by condition), to conclude that $BC$ is the external bisector of $\angle AMB $.

It suffices to prove that $ \frac{BP}{PM}=\frac{AB}{AM} $ It s clear that $AMB \sim NMC\implies \widehat{AMB} = \widehat{NMC} $ then $\widehat{PMB} = \widehat{AMC}$ thus $PMB \sim AMC \implies \frac{BP}{PM}=\frac{AC}{AM} $but $AB=AC$ therfore follow the result .

sasanineq wrote: let AT is thr line such that AT and MN are parallel.then use the lemma: in triangle $ABC$,if AD is internaly bisector of angle BAC then we have :BD/CD=AB/AC let $AT$ is the line such that $AT$ and $MN$ are parallel. then use the lemma in triangle ABC, if $AD$ is the internal bisector of angle $BAC$ then we have $\frac{BD}{CD}=\frac{AB}{AC}$

let $D$ the intersection of $AB$ and the perpendicular to $BC$ at $M$. clearly, $AMB\sim NMC \rightarrow AMD=MDP$. i.e $MD$ is the angle bisector of $AMP$, thus $(B,D;A,P)=-1 \rightarrow AB/PB=AD/PD=AM/MP$ and we r done by the angle bisector theorem.