We easy see that area of ABC=ABP+BCP+ACP and we prove that

From $A$ introduce $AH \perp BC$. In $\triangle ABC$, $\angle ABE=60^o$, implies $BE= \frac 12 AB= \frac 12 a$. Hence, applying Pythagoras' Theorem, we have $AE = \sqrt{ AB^2-BE^2}= \frac{ \sqrt 3 a}{2}$. It follows that $[ABC]= AE \cdot BC \div 2= \frac{ \sqrt 3 a^2}{4}$. We have $[APB]+[APC]+[BPC]=[ABC]$, $[APB]+[APC]+[BPC]= \frac 12a (|PD|+|PF|+|PE|)$. Therefore $a(|PD|+|PF|+|PE|)= \frac{ \sqrt 3 a^2}{2}$ or $\dfrac{|PD|+|PF|+|PE|}{3a}= \frac{ \sqrt 3}{6}$.

By Viviani’s Theorem, $$|PD|+|PE|+|PF| = \frac{a\sqrt{3}}{2}$$ Therefore, $$ \frac{|PD|+|PE|+|PF|}{3a} = \frac{\frac{ a\sqrt{3}}{2} } {3a} = \frac{a\sqrt{3}}{6a} = \boxed{\frac{\sqrt{3}}{6}.} $$