As $\angle ABC = \angle AP'C$ (see figure), it suffices to show that the quadrilateral $APCP'$ is a square if and only if $\angle AP'C = 90^\circ$.

Since $\angle ABP = \angle CBP$, we have $AP = CP$. Thus $PP'$ is the perpendicular bisector of the side $AC$. Hence the line $PP'$ passes through the circumcentre of the triangle $ABC$, i.e., the chord $PP'$ is a diameter. By Thales' theorem, $\angle PAP' = \angle PCP' = 90^\circ$.

If $\angle AP'C = 90^\circ$ then also $\angle APC = 180^\circ - 90^\circ = 90^\circ$. Thus the quadrilateral $APCP'$ is a rectangle and, by equality of adjacent sides $AP$ and $CP$, a square. On the other hand, if $APCP'$ is a square then obviously $\angle AP'C = 90^\circ$. This completes the proof.

Since $\angle ABC = \angle AP'C$, it suffices to show that the quadrilateral $APCP'$ is a square if and only if $\angle AP'C = 90^\circ$.

As $\angle ABP = \angle CBP$, we have $AP = CP$. Thus $PP'$ bisects the line segment $AC$.

If $\angle AP'C = 90^\circ$ then, by Thales' theorem, $AC$ is the diameter of the circumcircle of the triangle $ABC$. Thus the chord $PP'$ bisecting $AC$ passes through the circumcentre of the triangle $ABC$, implying that $AC$ bisects the chord $PP'$. As $PP' \perp AC$, the diagonals of the quadrilateral $APCP'$ are perpendicular and bisect each other. Hence the quadrilateral $APCP'$ is a rhombus and, because of $\angle AP'C = 90^\circ$, a square. On the other hand, if $APCP'$ is a square then obviously $\angle AP'C = 90^\circ$. This completes the proof.