Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Since $\angle P B A+\angle P C A+\angle P B C+\angle P C B=\beta+\gamma$, the condition from the problem statement is equivalent to $\angle P B C+\angle P C B=(\beta+\gamma) / 2$, i. e. $\angle B P C=90^{\circ}+\alpha / 2$. On the other hand $\angle B I C=180^{\circ}-(\beta+\gamma) / 2=90^{\circ}+\alpha / 2$. Hence $\angle B P C=\angle B I C$, and since $P$ and $I$ are on the same side of $B C$, the points $B, C, I$ and $P$ are concyclic. In other words, $P$ lies on the circumcircle $\omega$ of triangle $B C I$. ! Let $\Omega$ be the circumcircle of triangle $A B C$. It is a well-known fact that the centre of $\omega$ is the midpoint $M$ of the $\operatorname{arc} B C$ of $\Omega$. This is also the point where the angle bisector $A I$ intersects $\Omega$. From triangle $A P M$ we have $$A P+P M \geq A M=A I+I M=A I+P M \text {. }$$ Therefore $A P \geq A I$. Equality holds if and only if $P$ lies on the line segment $A I$, which occurs if and only if $P=I$.