Without loss of generality, $AC>AB$. Let $L$ is intersection of $BC$ and $AP$. Easy to see that$\angle PLC>90^{\circ}\Rightarrow PC>PB\Rightarrow PN>PM\Rightarrow AM+PN>AN+PM$, contradiction.

Approach by luisgeometria here. Here it is roughly translated: $ ANPM$ is circumscriptible in a circle $ \Longleftrightarrow$ $ AM + PN = AN + PM$ $ \Longrightarrow$ $ AB + \frac {_2}{^3}BM = AC + \frac {_2}{^3}CN$ If $ L$ is the midpoint of $ BC$ and $ Q$ is the reflection of $ P$ with respect to $ L$ we have $ AB - AC = \frac {_2}{^3}(CN - BM) = QB - QC^$. This means that $ Q$ lies on the same arm of the hyperbola $ \mathcal{H}$ with foci $ B,C$ passing through $ A$. From the symmetry of the hyperbola with respect to its centre $L$ , any line through $ L$ and cutting though $A$ has to cut $\mathcal{H}$ for the second time in it's other arm, which is a contradiction since that means $LA$ has cut twice in one arm of $\mathcal{H}$ unless it is degenerate in the bisector of $BC \Longrightarrow$ $ \triangle ABC$ has to be isosceles with apex $A$.

Bars wrote: $\angle PLC>90^{\circ}\Rightarrow PC>PB$ Can you please tell why this is so?

Law of cosines for triangles $PLB$ and $PLC$