jmc

Since $\angle QBP$ is created by trisecting $\angle ABC$, we have that $m\angle ABC=3\cdot 14=42$ degrees. Thus, the measure of $\angle ACB=180-42-39=99$ degrees. By the trisector information given, we have that $\angle PCB=99/3=33$ degrees and $\angle PBC=14$ degrees. Looking just at triangle $PBC$, the measure of $\angle BPC=180-14-33=\boxed{133}$ degrees.