Let $\Gamma_{\mathrm{a}}, \Gamma_{\mathrm{b}}$ and $\Gamma_{\mathrm{c}}$ be the circles centered at $P$ and passing through $A$, $B$, and $C$ respectively. Let $C^{\bullet}$ be the symmetric point of $C$ with respect to $(AP)$. Note that $C^{\bullet}$ is the only point on $\Gamma_{\mathrm{c}}$, other than $C$ itself, such that $A C = A C^{\bullet}$.
Now, let $r$ be the rotation centered at $\mathcal{A}$ with an angle of $60^{\circ}$ in the clockwise direction. Without loss of generality, we assume that $C = r(B)$.
Then, for any point $B^{\prime}$ on $\Gamma_{b}$, we associate the point $C^{\prime} = r(B^{\prime})$. The triangle $A B^{\prime} C^{\prime}$ is equilateral, and the hypothesis of the problem states that for any $B^{\prime}$, the point $C^{\prime}$ can only belong to $\Gamma_{c}$ if it is equal to $C$ or $C^{\bullet}$.
As $B^{\prime}$ describes $\Gamma_{b}$, the point $C^{\prime}$ describes the circle $r(\Gamma_{b})$, centered at $r(P)$ and with radius $PB$. This circle cannot contain both points $C$ and $C^{\bullet}$, since then its center would lie on the perpendicular bisector $(AP)$ of $[C^{\bullet}]$, which is not the case for $r(P)$.
Since $r(\Gamma_{b})$ already contains $C$, it cannot contain any other point of $\Gamma_{c}$. Given that $r(P)A = PA < PB$, we know that $A$ is located inside $r(\Gamma_{b})$, which is therefore internally tangent to $\Gamma_{c}$ at the point $C$.
Since $PB = r(P)C < PC$, it follows that the points $C, r(P)$, and $P$ are collinear in that order. This implies that $PC = r(P)C + r(P)P = PB + PA$, or equivalently, $CP \cdot AB = BP \cdot AC + AP \cdot BC$. The Ptolemy's theorem then indicates that the points $A, B, C$, and $P$ are concyclic, which concludes the proof.
!