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
$$|PA| \cdot |PD| = |PB| \cdot |PC| \quad (1)$$

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
$$|PA| = |PB| + |PC| \quad (2)$$
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
$$\frac{1}{|PD|} = \frac{1}{|PB|} + \frac{1}{|PC|}$$