$A{D^2} - P{D^2} + 3B{D^2} - 3P{D^2} = A{C^2} - Q{C^2} + 3B{C^2} - 3Q{C^2}\mathop \Leftrightarrow \limits^{AD = AC} $ $3(B{C^2} - B{D^2}) = 4(Q{C^2} - P{D^2}) \Leftrightarrow \frac{{Q{C^2} - P{D^2}}}{{B{C^2} - B{D^2}}} = \frac{3}{4}$ But, $\frac{{B{D^2}}}{{B{C^2}}} = \frac{{P{D^2}}}{{Q{C^2}}} \Leftrightarrow \frac{{B{C^2} - B{D^2}}}{{B{C^2}}} = \frac{{Q{C^2} - P{D^2}}}{{Q{C^2}}} \Leftrightarrow \sin B = \frac{{QC}}{{BC}} = \frac{{\sqrt 3 }}{2} \Leftrightarrow $ $\boxed{\angle B=60^\circ}$