You don’t need to know the side lengths

Let $D$ be the reflection of $A$ accross the line $BP$ . Then $\angle APB=180-(10+20)=150$ . Since $D$ is a reflection of $A$ accross the line $BP$ , $\angle PBD=\angle PBA=20$ , $\angle BPD=\angle BPA=150$ and $\angle PDB=\angle PAB=10$ . Therefore, $\angle APD=360-\angle BPD-\angle BPA=360-150-150=60$ . Since $AP=PD$ , triangle $APD$ is an equilateral triangle. Since $\angle BAC=50$ and $\angle ABD=40$ , $BD$ is perpendicular to $AC$ . Let $E$ be the intersection of $BD$ and $AC$ . $\angle PED=180-\angle CED=180-(90-30)=120$ . And so $\angle PED+\angle DAP=180$ . We conclude that quadrilateral $APED$ is a cyclic, and therefore $\angle DEA=\angle DPA=60$ . Since $\angle AED=\angle CED=60$ and $BD$ is perpendicular to $AC$ , we conclude that $AB=BC$ . That means that $\angle ACB=50$ . Finally. $\angle PCB=50-30=20$ .

Let $\angle PCB = \theta$ . We note that $\angle CPA = 180^\circ - 30^\circ - 40^\circ = 110^\circ$ and $\angle BPA = 180^\circ - 10^\circ - 20^\circ = 150^\circ$ , therefore, $\angle BPC = 360^\circ - 110^\circ - 150^\circ = 100^\circ$ and $\angle CBP = 180^\circ - 100^\circ - \theta = 80^\circ - \theta$ . Now we can find $\theta$ using sine rule on $\triangle CBP$ as follows:

$\dfrac {\sin \theta}{BP} = \dfrac {\sin (80^\circ - \theta)}{CP}$

We only need to find $BP$ and $CP$ . Let $AC = 1$ . By sine rule:

$\begin{aligned} \frac {CP}{\sin 40^\circ} & = \frac 1{\sin 110^\circ} \\ \implies CP & = \frac {\sin 40^\circ}{\sin 110^\circ} \\ \frac {BP}{\sin 10^\circ} & = \frac {\color{#3D99F6}AP}{\sin 20^\circ} & \small \color{#3D99F6} \text{Note that } AP = \frac {\sin 30^\circ}{\sin 110^\circ} \\ \implies BP & = \frac {{\color{#3D99F6}\sin 30^\circ}\sin 10^\circ}{{\color{#3D99F6}\sin 110^\circ}\sin 20^\circ} \end{aligned}$

Now we have:

$\begin{aligned} \frac {\sin \theta}{BP} & = \frac {\sin (80^\circ - \theta)}{CP} \\ \sin \theta & = \frac {BP}{CP} \sin (80^\circ - \theta) \\ \sin \theta & = \frac {\sin 30^\circ \sin 10^\circ}{\sin 110^\circ \sin 20^\circ} \times \frac {\sin 110^\circ}{\sin 40^\circ} (\sin 80^\circ \cos \theta - \cos 80^\circ \sin \theta) \\ \sin \theta & = \frac {\sin 30^\circ \sin 10^\circ}{\sin 110^\circ \sin 20^\circ} \times \frac {\sin 110^\circ}{\sin 40^\circ} (\sin 80^\circ \cos \theta - \cos 80^\circ \sin \theta) \\ \sin \theta & = \frac {\cos \theta}{4 \sin 40^\circ} - \frac {\tan 10^\circ \sin \theta}{4 \sin 40^\circ} \\ 4 \sin 40^\circ \sin \theta & = \cos \theta - \tan 10^\circ \sin \theta \\ \implies \tan \theta & = \frac 1{4\sin 40^\circ + \tan 10^\circ} \\ & = \frac {\color{#3D99F6}\sin 20^\circ}{4{\color{#3D99F6}\sin 20^\circ}\sin 40^\circ + {\color{#3D99F6}\sin 20^\circ}\tan 10^\circ} \\ & = \frac {\sin 20^\circ}{2(\cos 20^\circ - \cos 60^\circ) + 2\sin 10^\circ \cos 10^\circ \times \dfrac {\sin 10^\circ}{\cos 10^\circ}} \\ & = \frac {\sin 20^\circ}{2 \cos 20^\circ - 1 + 2\sin^2 10^\circ} \\ & = \frac {\sin 20^\circ}{2 \cos 20^\circ - \cos 20^\circ} \\ & = \tan 20^\circ \\ \implies \theta & = \boxed{20^\circ} \end{aligned}$

Angles easy to obtain are listed in image. Set $CP=1$ . Law of sines in $APC$ is $a=\sin(30)\frac{1}{\sin(40)}$

Law of sines in $APB$ gives $b=\sin(10)\frac{a}{\sin(20)}=\frac{\sin(10)}{\sin(20)} \frac{\sin(30)}{\sin(40)}=0.39493$

Law of cosines in $CPB$ gives us $c=\sqrt{b^2+1-2b\cos(100)}=1.13716$

Law of sines in $CPB$ finally results in $\angle BCP=\arccos(\frac{1+c^2-b^2}{2c})=20^\circ$