Notice that $BACQ$ is cyclic. Now by Ptolemy's we get $PC \cdot BA + BP \cdot CA = PA \cdot BC$ since $ABC$ is equilateral we get $PA=PC+PB$. Now notice that triangles $APB$ and $PQC$ are similar, so $\frac{PQ}{PC}=\frac{PB}{PA}=\frac{PB}{PB+PC}$. Finally we get $PQ=\frac{PB\cdot PC}{PB+PC}$ so $\frac{1}{PQ}=\frac{PB+PC}{PB\cdot PC}=\frac{1}{PC}+\frac{1}{PB}$

Since $QCAB$ is cyclic we have $BC\cdot AP=AC\cdot BP+AB\cdot PC$ by Ptolemy's. Moreover $\triangle APB\sim\triangle PQC$ so $\frac{PQ}{PC}=\frac{PB}{PA}$. Also, note that $AP=BP+CP$, since $\triangle ABC$ is equilateral. Hence \[\frac{PQ}{PC}=\frac{PB}{PB+PC}\implies PQ=\frac{PB\cdot PC}{PB+PC}\implies \frac{1}{PQ}=\frac{PB+PC}{PB\cdot PC}\] as desired.

PP. Let $ABC$ be an $A$-isosceles triangle, i.e. $AB=AC$ . For a point $Q\in (BC)$ denote the second intersection $P$ of the line $AQ$ with the circumcircle of $\triangle ABC$ . Prove that $\frac {1}{PB}+\frac {1}{PC}=\frac {k}{PQ}$ , where $k=\frac {BC}{AB}$ . Prooof 1 (without the Ptolemy's theorem). $\left\{\begin{array}{ccc} \triangle PBQ\sim \triangle CAQ & \implies & \frac {PQ}{PB}=\frac {CQ}{CA}\\\\ \triangle PCQ\sim \triangle BAQ & \implies & \frac {PQ}{PC}=\frac {BQ}{BA}\end{array}\right\|\ \bigoplus\ \implies$ $PQ\cdot\left(\frac {1}{PB}+\frac {1}{PC}\right)=\frac{BC}{AB}\implies$ $\boxed{\frac {1}{PB}+\frac {1}{PC}=\frac {k}{PQ}\ ,\ \mathrm{where}\ k=\frac {BC}{AB}}$ . Prooof 2 (trigonometric). Denote $\left\{\begin{array}{c} m\left(\widehat{QAB}\right)=x\\\ m\left(\widehat{QAC}\right)=y\\\ m\left(\widehat{AQB}\right)=\phi\end{array}\right\|$ . Observe that $B+y=\phi$ , $B+\phi =180^{\circ}-x$ and $BC=2\cdot AB\cos B$ . Thus, $\frac {1}{PB}+\frac {1}{PC}=$ $\frac {k}{PQ}\iff$ $\frac {PQ}{PB}+\frac {PQ}{PC}=$ $\frac {BC}{AB}\iff$ $\frac {\sin y}{\sin \phi}+\frac {\sin x}{\sin\phi}=$ $2\cos B\iff$ $\sin x+\sin y=2\cos B\sin\phi =$ $\sin (B+\phi)+\sin (\phi-B)$ , what is truly. Extension. Let $\triangle ABC$ and for $Q\in (BC)$ denote the second intersection $P$ of the line $AQ$ with the circumcircle of $\triangle ABC$ . Prove that $\boxed{\frac {AC}{PB}+\frac {AB}{PC}=\frac {BC}{PQ}}$ . Proof. $\left\{\begin{array}{ccc} \triangle ABQ\sim\triangle CPQ & \implies & \frac {AB}{CP}=\frac {BQ}{PQ}\\\\ \triangle ACQ\sim\triangle BPQ & \implies & \frac {AC}{BP}=\frac {CQ}{PQ}\end{array}\right\|\ \bigoplus\ \implies\ \frac {AC}{PB}+$ $\frac {AB}{PC}=\frac {BC}{PQ}$ . See PP2 from here.

Everyone here has their cool and unique Ptolemy’s and trig solutions, and I’m stuck here with my stupid comp solution ;-; First, WLOG, assume the side length of $\triangle{ABC}$ is $1$. Then, let $BQ=x$ and $CQ=1-x$. Using LOC on $ABQ$ we get that $AQ$ is $\sqrt{x^2-x+1}$. Because of PoP, $AQ \cdot PQ=BQ \cdot CQ$, and we get that $PQ=\frac{x(1-x)}{\sqrt{x^2-x+1}}$, or $\frac{1}{PQ}=\frac{\sqrt{x^2-x+1}}{x(1-x)}$. Next, because of the inscribed angle theorem, $\frac{\sqrt{x^2-x+1}}{1-x}=\frac{1}{PC}$ Similarly, we get that $\frac{(\sqrt{x^2-x+1})}{x}=\frac{1}{PB}$. Adding the fractions up, we can in fact conclude that $\boxed{\frac{1}{PB}+\frac{1}{PC}=\frac{1}{PQ}}$. This took me way too long to type up