Drawing and Measuring is Prohibited

Extend $\overline{CP}$ and intersects $\overline{AB}$ at $E$ . Construct $\overline{EF}\perp \overline{BP}$ intersecting $\overline{BC}$ at $F$ .

$\overline{CE}$ bisects $\angle ACB$ as $m\angle BCP=\frac{1}{2}m\angle BCA$ . $\triangle ACE\cong \triangle FCE$ ( $A.S.A.$ ), then $\overline{AC}\cong \overline{FC}$ .

Notice $m\angle AEC=60°$ , thus $m\angle FEC=60°$ , then $m\angle BEF=60°=m\angle PEF$ . $\overline{EF}$ is both the angle bisector and the height of $\triangle BEP$ , indicating that $\triangle BEP$ is isosceles. Therefore, $\triangle BFP$ is also isosceles. $m\angle PFC=2m\angle PBF=20°$ .

$\triangle ACP\cong \triangle FCP$ ( $S.A.S.$ ), then $m\angle CAP=20°$ .

$\text{Since angle ABC= angle ACB, both must be=(180 - 100)/2=40. so angle ABP=30, and angle ACP=20.}\\ Let \ AB=AC=x, \ AP=y,\ \ \ \angle\ APB=M,\ \ \angle\ APC=N. \ \ \therefore \ M+N=180-10 - 20=150.\ \ \implies\ M=150 - N.\\ \text{Applying Sin Law in the form } \dfrac{SinP}{SinQ}= \dfrac p q, \ to\ \Delta s\ ABP\ and\ ACP.\\ \dfrac{Sin30}{SinM}= \dfrac y x= \dfrac{Sin20}{SinN},\ \ \\ \dfrac{Sin30}{SinM}= \dfrac{Sin20}{SinN}, \implies\ \dfrac{Sin(150- N)}{SinN}=\dfrac{Sin30}{sin20}.\\ Expanding \ Sin(150 - N)\ and\ \ simplifying, \ we\ get - CotN + Cot30=\dfrac 1 {Sin20}.\\ \text{Inserting values we finally get N=140.}\ \ \therefore\ \angle\ PAC=180 - 20 - 140=20^o.$

Triangle CAP is isosceles so angle CAP=20 degrees

By the way, I am Chinese.