It suffices to show that the incircle of $\triangle{AMN}$ is tangent to $KP$. By 2x homothety at $A$, it suffices to show that the tangent from the reflection of $A$ over $K$ to the incircle of $\triangle{ABC}$ is the reflection of $AB$ over the perpendicular from incenter of $\triangle{ABC}$ to $AC$, which this is obvious by symmetry

I'll show that AK+MP=AM+KP. Assume $$\alpha=\frac{NK}{NH}$$. We need to prove that $$p-a+\frac{a}{2}.(1-\alpha)=\frac{c}{2}+\frac{c}{2}.\alpha$$. Which is equivalent that we must prove that $$\alpha=\frac{b}{a+c}$$. $$\alpha=\frac{NK}{NH}=\frac{b}{a+c}\Leftrightarrow\frac{b}{a+c}=\frac{a/2-c/2}{b/2-AH}\Leftrightarrow AH=\frac{b^2-a^2+c^2}{2b}$$.Which is true :$$power of A to (BHC)=AH.b=m_a^2-a^2/4=\frac{b^2-a^2+c^2}{2}$$