Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ, \angle{BAP}=\angle{BCP} = 20^\circ$. Show $\triangle{ABC}$ is an isosceles Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ$, $\angle{BAP}=\angle{BCP} = 20^\circ$. Show $\triangle{ABC}$ is an isosceles and find $\angle{CPA}$.
extend $AP$, and tried law of sines; also googled and searched on stackexchange, no clues.
This seems to be one of the hard IMO problems https://davidaltizio.web.illinois.edu/CollectionOfGeometryProblems.pdf or Sharygin geometry problems https://www.isinj.com/mt-aime/Problems%20in%20Plane%20Geometry%20-%20Sharygin%20(MIR,1982).pdf and no good solution yet.
As shown by answers, the original problem doesn't give enough conditions to prove what's asked. The missing part is $\angle PBC=30^\circ$. Then one can make a mirror point $P'$ of $P$ over $BC$, connect $AP'$, $AB=PC$, $\Delta BPC$ is equilateral, and $BAP$ and $P'AP$ are congruent, $\Delta AP'C$ is equilateral, so $AB=PC=AP'=P'C=AC$.
 A: These conditions do not uniquely determine a triangle. By the inscibed angle theorem, $C$ could be anywhere on the green circle such that $P$ is still in $\triangle ABC$:
$\hspace{3cm}$ 
A: The stated problem

Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ$, $\angle{BAP}=\angle{BCP} = 20^\circ$. Show $\triangle{ABC}$ is an isosceles and find $\angle{CPA}$.

has missing data since $\triangle{ABC}$ is not necessarily an isosceles triangle. See the end of this answer for the details.
Then, you added

As shown by answers, the original problem doesn't give enough conditions to prove what's asked. The missing part is $\angle PBC=30^\circ$. Then one can make a mirror point $P'$ of $P$ over $BC$, connect $AP'$, $AB=PC$, $\Delta BPC$ is equilateral, and $BAP$ and $P'AP$ are congruent, $\Delta AP'C$ is equilateral, so $AB=PC=AP'=P'C=AC$.

It seems that the problem you want to solve is

Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ$, $\angle{BAP}=\angle{BCP} = 20^\circ$ and $\color{red}{\angle{PBC}=30^\circ}$. Show $\triangle{ABC}$ is an isosceles and find $\angle{CPA}$.

It seems that your proof has an error (or a typo) since $\triangle{BPC}$ is not an equilateral triangle.
In the following, I'll show my solution which uses your idea taking $P'$, which I think is a nice idea.

Since $BP=BP'$ and $\angle{PBP'}=60^\circ$, $\triangle{PBP'}$ is an equilateral triangle. Let $D$ be the intersection point of $AP$ with $BP'$. Since $\angle{BPD}=30^\circ$, one can see that $\angle{BDP}=90^\circ$ with $DB=DP'$. Since $\triangle{ADB}\equiv \triangle{ADP'}$, one can see that $AB=AP'$ and $\angle{AP'P}=10^\circ$. It follows that $\triangle{ABP'}\equiv\triangle{CPP'}$ since $BP'=PP', \angle{ABP'}=\angle{CPP'}$ and $\angle{AP'B}=\angle{CP'P}$. So, it follows from $AP'=AB=CP=CP'$ and $\angle{AP'C}=60^\circ$ that $\triangle{AP'C}$ is an equilateral triangle. Therefore, one finally gets $AB=CP=CP'=AC$, so $\triangle{ABC}$ is an isosceles triangle.
So, $\angle{CPA}=180^\circ-\angle{CPP'}-\angle{P'PD}=180^\circ-70^\circ-30^\circ=80^\circ$.

The stated problem has missing data since $\triangle{ABC}$ is not necessarily an isosceles triangle.
If $\angle{ABP}=10^\circ, \angle{BAP}=\angle{BCP} = 20^\circ,\angle{PBC}=20^\circ$ and $\angle{PCA}=30^\circ$, then $\triangle{ABC}$ is not an isosceles triangle.
One has
$$\begin{align}\frac{\sin\angle{PBA}}{\sin\angle{PAB}}\cdot\frac{\sin\angle{PCB}}{\sin\angle{PBC}}\cdot\frac{\sin\angle{PAC}}{\sin\angle{PCA}}
&=\frac{\sin 10^\circ}{\sin 20^\circ}\cdot\frac{\sin 20^\circ}{\sin 20^\circ}\cdot\frac{\sin 80^\circ}{\sin 30^\circ}
\\\\&=\frac{2\sin 10^\circ\sin 80^\circ}{\sin 20^\circ}
\\\\&=\frac{\cos(10^\circ-80^\circ)-\cos(10^\circ+80^\circ)}{\sin 20^\circ}
\\\\&=\frac{\cos 70^\circ}{\sin 20^\circ}
\\\\&=1\end{align}$$
so by the trigonometric form of Ceva's theorem, one can say that such $P$ and $\triangle{ABC}$ exist.
A: initial conditions: Let $P$ be a point inside $△ABC$ such that $∠ABP=10^\circ$, $∠BAP=∠BCP=20^\circ$
if $C\in  R=λ_{A}∪λ_{B}∪l$ then the triangle ABC is an isosceles.
$\begin{array}{} λ_{A}(A,1) & λ_{B}(B,1) & (l) \text{ perpendicular bisector of the line segment AB.} \end{array}$
If $C$ belongs to the intersection of $R$ with the circle $\lambda'$ then we have the solution satisfying the conditions.
($λ'$ is the reflection of $λ(O,1)$ with respect to line $PB$). $\lambda'$ is locus of C such that $\angle{BCP}={20^\circ}$. Two solutions:$C$ and $C_{1}$.  $O$ is the circumcenter of the triangle $ABP$. $O=(\frac{1}{2},-\frac{\sqrt{3}}{2})$.
straight line $AP$ through $A(0,0)$ with slope equal to $tan (\angle{20}^\circ)$ and $BP$ through $B(1,0)$ with slope equal to $-tan (\angle{10}^\circ)$. ($P=AP∩PB$) ($(l)$ $x=\frac{1}{2}$)
$h=tan\left(  \frac{π}{18}  \right)$
$\begin{array}{} x_{P}=\frac{h^2-1}{h^2-3} & y_{P}=\frac{2h}{3-h^2}  \end{array}$
$\begin{array}{} x_{O´}=\frac{3h^2-5}{2(h^2-3)} & y_{O'}=\frac{\sqrt{3}(h^2-3)-4h}{2(h^2-3)}  \end{array}$
$C=\left( \frac{1}{2},\frac{\sqrt{3}(h^2-3)-4(h+\sqrt{2-h^2})}{2(h^2-3)}  \right)$
$b^2=\frac{[h(8-2\sqrt{3}h)+6\sqrt{3}]\sqrt{2-h^2}+h^3(h-2\sqrt{3})-6h(h-\sqrt{3})+17}{(h^2-3)^2}$
$t^2=\frac{h^2-5-2\sqrt{3}\sqrt{2-h^2}}{h^2-3}$
$d=\frac{h^2+1}{h^2-3}$
$cos(θ)=\frac{-b^2+d^2+t^2}{2dt}$
$θ=cos^{-1}\left( \frac{h^3(h+2\sqrt{3})-2h\left( 3\sqrt{3}+4\sqrt{2-h^2}-1  \right) }{2(h^2+1)\sqrt{15+h^2(h^2-8)-2(h^2-3)\sqrt{6-3h^2}}}  \right)$
$\theta≅115.476252^\circ$
where: $b=AC$, $t=CP$, $d=AP$,  $\theta=\angle{APC}$
$θ_{2}=∠(APC_{1})=80°$

