Irska

Because $\angle PAC=\angle PBC$, $\angle APC=\angle ABC=60^{\circ }$ and $\angle BPA=\angle BCA=60^{\circ }$ the triangles $APC$ and $BPD$ are similar. Thus $\frac{|PA|}{|PB|}=\frac{|PC|}{|PD|}$, so $$|PA|\cdot |PD|=|PB|\cdot |PC|(1)$$ Since $ABPC$ is a cyclic quadrilateral, $|PA|\cdot |BC|=|PB|\cdot |AC|+|PC|\cdot |AB|$. But the triangle $ABC$ is equilateral so it follows that $$|PA|=|PB|+|PC|(2)$$ From the two equations (1) and (2), it follows that $$|PB|\cdot |PC|=|PD|(|PB|+|PC|).$$ Now dividing by the product $|PB|\cdot |PD|\cdot |PC|$, we get the desired equality $$\frac{1}{|PD|}=\frac{1}{|PB|}+\frac{1}{|PC|}$$