If $P$ is not within the angle formed by $AB$ and $BC$ that is part of triangle $ABC$, $\angle ABC = 60^{\circ}$ so $\triangle ABC$ is equilateral. Then, $\triangle BPC$ is isosceles so we get $80^{\circ}$. second case wrong

As in #2, if $P$ and $A$ are in different half-planes w.r.t the line $BC$, $\angle{ABC}=\angle{ABP}-\angle{CBP}=60^\circ$, which shows that $\triangle {ABC}$ is equilateral. Then $BP=AC=BC$, so $\triangle {BPC}$ is isosceles, with $\angle{CBP} = 20^\circ$. We obtain $\angle{BCP}=80^\circ$. Instead, if $P$ and $A$ are in the same half-plane w.r.t the line $BC$, it can be easily seen that ray $[BP$ must lie in the interior of the angle $\angle{ABC}$. Therefore, $\angle{ABC}=\angle{ABP}+\angle{CBP}=100^\circ$, whence 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 to be $90^\circ$; if one is not familiar with said configuration, one may take the reflection $E'$ of $E$ w.r.t $C$ and effortlessly prove that $\triangle{EPE'}$ is equilateral in order to reach the same conclusion in regards to $\angle{PCE}$. Finally, since $\angle{PCA}+\angle{BCA}=90^\circ+40^\circ$, we conclude that $\angle{BCP}=130^\circ$.

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