Suppose that $PA=X$ , $PB=Y$ , $PC=Z$. Than we rotate the $\triangle{APC}$ around point $B$ with $60^\circ$ counterclockwise. Now we have new triangle ($\triangle{KQA}$). $KA=KB$ , $\widehat{KAB}$ $\Rightarrow$$\triangle{KAB}$ is equilateral. $QK=PA=X$ , $QA=PC=Z$ , $QB=PB=Y$ $\Rightarrow$ $\triangle{PBQ}$ is equilateral. We deduce $\triangle{PAQ}$ is a triangle with edges $X$ , $Y$ , $Z$.According to assume we know that $\widehat{AQP}=$ $90^\circ$. $\triangle{AQB}=\triangle{BPC}$$\Rightarrow$ $\widehat{BQA}=\widehat{BPC}=150$. Solution by Amirhossein Shahabi

law of sines on $\bigtriangleup APC$ yields $\angle PAC=45^{\circ}$ and then $\angle BPC = 150^{\circ}$ by simple angle chase