The solution given at the linked website is not clear and also wrong,
so I'll write a clearer and correct solution here. To really
understand this, you probably need to draw a figure (as I did when
writing this).
Let $J$ be the circumcenter of $\triangle CIB.$ The key part of this
problem is proving that $IJ$ is parallel to $AI.$ Since $JC,$
$JB,$ and $JI$ are radii of the circumcircle, their lengths are all
equal to the same value (that is, $|JC|=|JB|=|JI|=R_{CBI}$). Triangles
$JCI,$ $JBI,$ and $JCB$ are isoceles, so $\angle JCI=\angle JIC,$
$\angle JBI = \angle JIB,$ and $\angle JBC=\angle JCB.$ Using
the fact that $I$ is the intersection of the angle bisectors of
$\triangle ABC,$ the sum of the interior angles of quadrilateral $CIBJ$ is
$$
\begin{align}
& 2(C/2+\angle JCB) + 2(B/2+\angle JCB) + (\pi - 2\angle JCB) \\
& \qquad = C+B + 2\angle JCB + \pi \\
& \qquad = \pi-B-A + B + 2\angle JCB + \pi \\
& \qquad = 2\pi - A + 2\angle JCB \\
& \qquad = 2\pi
\end{align}
$$
so $\angle JCB = A/2$ and $\angle JIB=A/2+B/2,$ which is
$\pi - \angle AIB.$ So $AI$ is parallel to $IJ.$
Let $P=I.$ Then $\angle PBA=B/2,$ $\angle PCA = C/2,$
$\angle PBC=B/2,$ and $\angle PCB=C/2$ (so the condition of the problem
is satisfied). When $P=I,$ $\angle PBA=\angle PBC$ and $\angle PCA=\angle PCB.$ Moving $P$ from $I$ means setting $\angle PBA = B/2-s,$
$\angle PBC=B/2+s,$ $\angle PCA=C/2-t,$ and $\angle PCB=C/2+t$ for some
angle increments $s$ and $t.$
The condition
$\angle PBA + \angle PCA = \angle PBC + \angle PCB$ is now equivalent to
$s=-t.$ So when decreasing $\angle PBA$ by $s$ (for some $s\in (-B/2,B/2)$),
$\angle PCA$ must be increased by $s.$ This means
$\angle PBI = \angle PCI = s.$ So a classical geometry
theorem says that quadrilateral $BPIC$ is cyclic, meaning $P$ is on
the circumcircle of $\triangle CBI,$ meaning $|JP| = |JI|.$
Finally, the triangle inequality applied to $\triangle APJ$ shows that
$|AP|+|PJ| \geq |AJ|=|AI|+|IJ|.$ Since $|PJ| = |IJ|,$ $|AP|\geq |AI|$
(with equality only when $\triangle APJ$ is degenerate, that is, when
$P=I$).