smc

Let $b$ be the measure of $\angle B$. $△BQP$ is an isosceles triangle, so $\angle BPQ=b$ and $\angle BQP=180-2b$. $AB$ is a line, so $\angle AQP=2b$. Since $△QPA$ is isosceles as well, $\angle QAP=2b$ and $\angle QPA=180-4b$. $BC$ is a line, so $\angle APC=3b$. Since $△APC$ is isosceles as well, $\angle ACB=3b$. $△ABC$ is also isosceles, so $\angle BAC=3b$, so $\angle PAC=b$. The angles in a triangle add up to $180$ degrees, so $3b+3b+b=180$. Solving the equation yields $b=\frac{180}{7}$. Thus, $\angle B=25{\frac{5}{7}}^{\circ }$, so the answer is $\boxed{25\frac{5}{7}}$.