Let $(AB'C') \cap (CA'B') = D$. We have $\angle DC'B+ \angle DA'B = \angle DC'A + \angle ACB + \angle BAC - \angle DA'C = 180^{\circ}$, because $\angle DC'A - \angle DA'C = (180^{\circ}-\angle DB'A) - (180^{\circ} - \angle DB'C) = \angle AB'C=\angle ABC$. So, $D \in (BC'A')$ and we are done.

Attachments:

$$(180^{\circ}-3\measuredangle ACB)+(180^{\circ}-3\measuredangle BAC)+(180^{\circ}-3\measuredangle CBA)=0^{\circ}$$$$\measuredangle A'CB'+\measuredangle B'AC'+\measuredangle C'BA'=0^{\circ}\Rightarrow$$$$\measuredangle A'PB'+\measuredangle B'PC'+\measuredangle C'PA'=0^{\circ}$$

Let $(AB'C')$ and $(BA'C')$ meet at $T$. We have $$\angle B'TC'=3\alpha -180,\angle A'TC'=3\beta-180,\angle A'CB'=3\gamma$$which clearly sum to 180 so we are done.