Rotate $\triangle ABP$ around point $B$, $90^{\circ}$, counterclockwise to be $\triangle CP'B$ ($P'$ is the new point after rotate of $P$.) We will get that $\angle PBP'=\angle PBC+\angle CBP'=\angle PBC +\angle PBA=90^{\circ}$. So, $PP'=\sqrt{PB^2+P'B^2}=\sqrt{2^2+2^2}=2\sqrt{2}$ $\angle BP'P=45^{\circ}$ Consider $\triangle PP'C$ $3^2=2\sqrt{2}^2+1^2$ $PC^2=P'P^2+P'C^2$ $\angle PP'^C=90^{\circ}$ $\Rightarrow \angle BP'C=45^{\circ}+90^{\circ}=135^{\circ}$ Therefore, $\angle APB=\angle BP'C=135^{\circ}$ $\Box$

This problem also appeared in the Baltic Way 2011.

Rotating point $P$ around point $B$ counterclockwise by $90^{\circ}$, we get that: $PP'^2=PB^2+P'B^2=2^2+2^2=8 \Rightarrow PP'= \sqrt{8} =2 \sqrt{2}$ If we focus on triangle $\bigtriangleup PP'C$, we get that: $3^2=(2 \sqrt{2})^2+1^2 \Rightarrow PC^2=PP'^2+CP'^2 \Rightarrow \angle PP'C=90^{\circ}$ $PB=P'B \Rightarrow \angle BP'P=45^{\circ} \Rightarrow \angle BP'C=90^{\circ}+45^{\circ}=135^{\circ}$ $\angle APB= \angle BP'C \Rightarrow \angle APB=135^{\circ}$ $\blacksquare$