In $\triangle ABC, PA+PB+PC=AB+AC, PA$ extend to cross $BC$ at $D, E$ is on $PD$, let $PE=x,f(x)=x+EB+EC$, then $f(x)$ is mono increasing function.
I can write
$BE=\sqrt{AB^2+(AP+x)^2-2AB*(AP+x) \cos \alpha} $ $CE=\sqrt{AC^2+(AP+x)^2-2AC*(AP+x) \cos \beta} $
$f'(x)=\dfrac{AP+x-AB*\cos \alpha}{BE}+\dfrac{AP+x-AC*\cos \beta}{CE}+1$
But prove $f'>0$ will be mess. I try to use geometric method to prove $PE+BE+CE > BP+CP$ but failed. I had assume $PE+BE+CE = BP+CP$, then get $PD+BD+DC=BP+PC$, it looks trivial wrong but not easy to explain it. I think there should be a simple way to solve it.
EDIT1: I had go further for $f"$ which is lucky that $f">0 \iff 1-\cos^2 \alpha \ge 0$ so $f'>0 \iff CP(AC-CP)^2+2(AC-CP)(AB-BP)(BP+CP)+BP(AB-BP)^2>0$ (with $AP=AB+AC-BP-CP$).
now I am looking for the geo solution.