Have a point $H\in\overline{BC}$ such that $\overline{AH}\perp\overline{BC}.$ Let $\overline{PH}=a,$ then it's obviously that $\overline{AH}=a$ and $\overline{BH}=\sqrt{3}a,$ so $\overline{BP}=(\sqrt{3}-1)a$ and hence $\overline{CP}=2\overline{BP}=2(\sqrt{3}-1)a.$ Here we can get $\overline{CH}=\overline{CP}-\overline{PH}=(2\sqrt{3}-3)a.$ Now in $RT\bigtriangleup{ACH},$ with the ratio $\frac{\overline{CH}}{\overline{AH}}=\frac{2\sqrt{3}-3}{\sqrt{3}}=2-\sqrt{3},$ we can know $\angle{ACB}=\boxed{75^{\circ}}.$

There's a more elementary solution using only the Pythagorean Theorem (several times) and similarity. You can see it if you decompose some angles.