Consider a triangle $ABC$ and mark the exterior angle corresponding to $\angle B.$

Notice that $\angle B$ and $\angle PBC$ form a linear pair. Therefore, their measures add up to $180^{\circ }.$ Also, by the Triangle Angle Sum Theorem, the sum of the measures of the angles of $△ABC$ add up $180^{\circ }.$ $$\{\begin{array}{l}m\angle B+m\angle PBC=180^{\circ }\\ m\angle A+m\angle B+m\angle C=180^{\circ }\end{array}$$ Next, subtract the second equation from the first one. $$\begin{array}{rl}m\angle B+m\angle PBC& =180^{\circ }\\ {}^{-}m\angle A+m\angle B+m\angle C& =180^{\circ }\\ m\angle PBC-m\angle A-m\angle C& =0\end{array}$$ Finally, solve the last equation for $m\angle PBC$ to obtain the desired result.