Find the angle $x$ in this triangle

Let $D$ be the circumcenter of $\triangle APB$. Since $\angle BPA=150^\circ$, we have $\angle ADB = 360^\circ - 2\angle BPA = 60^\circ$, and since $DA=DB$ it follows that $\triangle ABD$ is equilateral. So, $AD=AB=AC$ and therefore $A$ is the circumcenter of $\triangle DBC$.

Now, $\angle PDB = 2\angle PAB = 40^\circ$ and $\angle CDB = \frac 12 \angle CAB = 40^\circ$. Hence $\angle PDB = \angle CDB$. It follows that $D,P,C$ are collinear. Now it is easy to find $\angle DPA = 90^\circ - \frac 12 \angle ADP = 90^\circ - \angle ABP = 80^\circ$. Thus $\angle APC = 180^\circ - \angle DPA = 100^\circ$.

Let $\angle ACP =z $. By trigonometric form of Ceva's theorem we have: $$\frac {\sin60^\circ}{\sin20^\circ}\frac {\sin10^\circ}{\sin40^\circ}\frac {\sin(50^\circ-z)}{\sin z}=1.\tag1 $$

Further we have the following property for product of sines: $$ \prod_{k=1}^{n-1}2\sin\frac{k\pi}n=n. $$ Particularly for $n=9$ it gives $$ (2^4\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ)^2=9 \implies\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\frac3{16}.\tag2 $$ Combining (1) and (2) we obtain: $$ \frac {\sin z}{\sin(50^\circ-z)}=\frac{\sin^2 60^\circ\sin10^\circ\sin80^\circ}{\frac3{16}} =4\sin10^\circ\cos10^\circ=\frac{\sin20^\circ}{\sin30^\circ}\implies z=20^\circ, $$ and finally $$x=180^\circ-60^\circ-z=100^\circ.$$