Cosines rule

The Cosine Rule for $\triangle APC$ and $\triangle BPC$, where $\angle ACP=\theta$, gives $$AP^2 = AC^2+PC^2-2AC\cdot PC \cos\theta$$ and $$PB^2= BC^2+PC^2-2BC\cdot PC \cos \theta$$ 

Hence $$\cos\theta=\frac{BC^2+PC^2-PB^2}{ 2BC\cdot PC}$$ and 

$$\cos\theta= \frac{AC^2+PC^2-AP^2}{ 2AC \cdot PC}$$

so we have 

$$\frac{BC^2+PC^2-PB^2}{ 2BC\cdot PC}=\frac{AC^2+PC^2-AP^2}{ 2AC \cdot PC}$$ from equating the right-hand sides of these equations.

Multiplying both sides by $\dfrac{2PC}{AB}$ gives

$$\frac{BC^2+PC^2-PB^2}{ BC\cdot AB}=\frac{AC^2+PC^2-AP^2}{ AC \cdot AB}$$ which we can rearrange to obtain 

$$\frac{AP^2}{AC.AB} +\frac{PC^2}{AB}\left(\frac{AC-BC}{BC\cdot AC}\right) = \frac{AC-BC}{AB}+ \frac{PB^2}{AB\cdot BC}$$ 

As $A$, $B$ and $C$ are in a straight line, $AB+BC=AC$, so $AC-BC=AB$. Therefore, we can simplify the previous equation to obtain $$\frac{AP^2}{ AB\cdot AC}+\frac{PC^2}{AC\cdot BC} = 1 + \frac{PB^2}{AB\cdot BC}$$ which is the required result.