Why is it Always P?

Rotate the entire figure by 60° around A anticlockwise. We obtain another two points. Let P' be the rotation of P by 60° and C' the rotation of C by 60°.

Now we know that $AP=AP'$ since it is just a rotation, and $\angle PAP'=60°$ . Thus $PP'=AP$ . But $PP'^{2}=AP^{2}=BP^{2}+CP^{2}=CP^{2}+CP'^{2}$ hence $\triangle PP'C$ is a right-angled triangle. Thus $\angle PCP'=\angle ACP+\angle ACP'=\angle ACP+\angle ABP=90°$ . Hence $\angle PBC+\angle PCB=180°-90°-60°=30°.$ and the result follows.

Due to symmetry, the extension of $AP$ will meet $BC$ at a right angle, $BP = CP$ and $\angle BAP = \angle CAP = 30^\circ$ .

Let $BP=CP=a$ . Since $AP^2=BP^2+CP^2 = 2a^2\quad \Rightarrow AP = \sqrt{2}a$

Let $\angle ABP = \theta$ . By Sine Rule, we have: $\space \dfrac {\sin {\theta}} {\sqrt{2}} = \dfrac {\sin {30^\circ}}{a} = \dfrac {1}{2a} \quad \Rightarrow \sin {\theta} = \dfrac {1}{\sqrt{2}} \quad \Rightarrow \theta = 45^\circ$ .

Therefore, $\space \angle PBC = \angle PCB = 60^\circ-45^\circ=15^\circ$ and $\space \angle BPC = 180^\circ - 2\times 15^\circ = \boxed {150^\circ}$

A pure logic proof:
If AP=BP=CP, angle BPC=120.
Legs PB and PC are shorter than PA.
So angle BPC>120.
But from the given selection, only 150>120.
So angle BPC=150.