You may assume your circle to be the unit circle in the $(x,y)$-plane and $P=(1,0)$. If the three parts have to have equal area then $A=\bigl(\cos(2\phi),\sin(2\phi)\bigr)$ and $B=\bigl(\cos(2\phi),-\sin(2\phi)\bigr)$ for some $\phi\in\ ]0,{\pi\over2}[\ $. Calculating the area over the segment $PA$ gives the condition $2\Bigl({\phi\over2}-{1\over2}\cos\phi\sin\phi\Bigr)={\pi\over3}\ ,$ or $f(\phi):=\phi-\cos\phi\sin\phi-{\pi\over 3}=0$. This equation has to be solved numerically. One finds $\phi\doteq1.30266$.