Explanation:

Let PB = $x$ and PC = $2x$
Drop a perpendicular from C on AP and name it as D.

In triangle CDP:
DCP = 30
CPD = 60
CDP = 90

Since, it is 30-60-90 triangle: CP = $2x$, PD = $x$ and CD = $\sqrt{3}x$

Now, let us join BD

In triangle PDB:
PD = PB = $x$
i.e. PDB = PBD = 30

Also, BD = $\sqrt{3}x$

In triangle ADB:
DAB = DBA = 15
ADB = 150
i.e. BD = AD = $\sqrt{3}x$

Lastly, In triangle ACD:
AD = CD = $\sqrt{3}x$
Therefore, CAD = ACD = $a$ (let)

i.e. $CAD + ACD + 90 = 180$
$2a = 90$
$a = 45$

Thus ACB = $a + 30 = 45 + 30 = 75$

Hence, option D