Let $ A',B',C'$ be the second intersections of rays $ AP,BP,CP$ with the circumcircle $ (O)$ of $ \triangle ABC.$ Angle chase gives $ \angle B'CP = \angle BPC - \angle BB'C = \angle BPC - \angle BAC = \alpha$ Similarly, $ \angle C'AP = \beta$ and $ \angle A'BP = \gamma.$ By sine law in the triangles $ \triangle PCB',\triangle PAC'$ and $ \triangle PBA'$ we obtain $ \frac {\sin\widehat{BAC}}{\sin \alpha} = \frac {PC}{PA} \ , \ \frac {\sin\widehat{CBA}}{\sin \beta} = \frac {PA}{PC'} \ , \ \frac {\sin\widehat{ACB}}{\sin \gamma} = \frac {PB}{PA'}$ From power of $ P$ WRT $ (O) \ , $ $PA \cdot PA' = PB \cdot PB' = PC \cdot PC'$ we get $ PA \cdot \frac {\sin\widehat{BAC}}{\sin \alpha} = PB \cdot \frac {\sin\widehat{CBA}}{\sin \beta} = PC \cdot \frac {\sin\widehat{ACB}}{\sin \gamma}$