1. **Case 1: $P$ and $A$ are in different half-planes with respect to the line $BC$**

- Given: $\angle ABP = 80^\circ$, $\angle CBP = 20^\circ$, and $AC = BP$.
- Since $\triangle ABC$ is isosceles with $AB = BC$, we have $\angle ABC = \angle ACB$.
- We know that $\angle ABP = 80^\circ$ and $\angle CBP = 20^\circ$.
- Therefore, $\angle ABC = \angle ABP - \angle CBP = 80^\circ - 20^\circ = 60^\circ$.
- This implies that $\triangle ABC$ is equilateral because $\angle ABC = 60^\circ$ and $AB = BC$.
- Since $\triangle ABC$ is equilateral, $AC = BC$ and $BP = AC = BC$.
- Thus, $\triangle BPC$ is isosceles with $BP = BC$.
- Given $\angle CBP = 20^\circ$, we can find $\angle BCP$ as follows:
$$\angle BCP = 180^\circ - \angle CBP - \angle BPC = 180^\circ - 20^\circ - 80^\circ = 80^\circ$$

2. **Case 2: $P$ and $A$ are in the same half-plane with respect to the line $BC$**

- Given: $\angle ABP = 80^\circ$, $\angle CBP = 20^\circ$, and $AC = BP$.
- Since $\triangle ABC$ is isosceles with $AB = BC$, we have $\angle ABC = \angle ACB$.
- We know that $\angle ABP = 80^\circ$ and $\angle CBP = 20^\circ$.
- Therefore, $\angle ABC = \angle ABP + \angle CBP = 80^\circ + 20^\circ = 100^\circ$.
- This implies that the angles of $\triangle ABC$ are $100^\circ, 40^\circ, 40^\circ$.
- Let $BP \cap AC = \{E\}$ and let $F$ be on the segment $[AC]$ such that lines $BE$ and $BF$ are isogonal (and thus isotomic since $BA = BC$).
- We get $CE = FA$. By symmetry, we have $BE = BF$.
- Moreover, $\angle EBF = 100^\circ - 2 \cdot 20^\circ = 60^\circ$.
- Consequently, $\triangle BEF$ is equilateral and $BE = EF$.
- It follows that $BP - BE = AC - EF \Longleftrightarrow PE = CE + FA$, which rewrites as $PE = 2 \cdot CE$.
- Now look at triangle $\triangle PCE$: $\angle CEP = 60^\circ$ and $PE = 2 \cdot CE$.
- This is a well-known configuration in which $\angle PCE$ turns out to be $90^\circ$.
- Finally, since $\angle PCA + \angle BCA = 90^\circ + 40^\circ$, we conclude that $\angle BCP = 130^\circ$.

The final answer is $\boxed{80^\circ \text{ or } 130^\circ}$