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Denote $\alpha = \angle BAC$, $\beta = \angle CBA$ and $\gamma = \angle ACB$, and let $O$ be the circumcentre of triangle $ABC$.

Point $P$ is on the circle circumscribed to triangle $ACC'$, so $\angle APC = \angle AC'C$ since they are subtended by $\text{AC}$. Because of the reflection, we have $\angle AC'C = 90^\circ - \angle BAC' = 90^\circ - \alpha$.
Analogously, by observing the circle circumscribed to triangle $ABB'$, we can conclude
$$\angle APB = \angle AB'B = 90^\circ - \alpha.$$
Therefore, $\angle CPB = \angle APC + \angle APB = 180^\circ - 2\alpha$.
Since $AC'PC$ is a cyclic quadrilateral, we have
$$\angle CPC' = 180^\circ - \angle CAB - \angle BAC' = 180^\circ - 2\alpha.$$
Therefore, $\angle CPB = \angle CPC'$, so points $P$, $B$ and $C'$ are collinear.
Let $A'$ be the point diametrically opposite point $A$ on the circle circumscribed to triangle $ACC'$. Since $AB$ is the bisector of segment $\text{CC'}$, point $A'$ lies on line $AB$. Therefore, $\angle A'BP = \angle ABC' = \beta$. By Thales' theorem we have $\angle APA' = 90^\circ$, so
$$\angle BPA' = 90^\circ - \angle APB = \alpha.$$
Therefore, $\angle AA'P = 180^\circ - \alpha - \beta = \gamma$ and $\angle A'AP = 90^\circ - \angle AA'P = 90^\circ - \gamma$, i.e. $\angle BAP = 90^\circ - \gamma$.
On the other hand, $\angle BOA = 2\gamma$ and $\angle BAO = 90^\circ - \gamma$, so we can conclude that $A$, $O$ and $P$ lie on the same line. This completes the proof.