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