The Cosine Rule for $\Delta APC$ and $\Delta BPC$, where $\angle ACP=\theta$, gives $\begin{eqnarray} AP^2 &= AC^2+PC^2-2AC.PC \cos\theta,\cr PB^2&= BC^2+PC^2-2BC.PC \cos \theta. \end{eqnarray}$

Hence $\begin{eqnarray} \frac{BC^2+PC^2-PB^2}{ 2BC.PC}&= \frac{AC^2+PC^2-AP^2}{ 2AC.PC} = \cos\theta. \end{eqnarray}$

Hence, multiplying both sides by $2PC/AB$, we find that $\begin{eqnarray} {AP^2\over AC.AB} +{PC^2\over AB}\left({AC-BC\over BC.AC}\right) &= {PB^2\over AB.BC} +{AC-BC\over AB}.\end{eqnarray}$ As $AB+BC=AC$, we get the result: $\begin{eqnarray} {AP^2\over AB.AC}+{PC^2\over AC.BC} &= 1 + {PB^2\over AB.BC}. \end{eqnarray}$