jmc

First, note that $P$ lies on the line $y-x\sqrt{3}+3=0.$

Solving for $x$ in $2y^2=2x+3,$ we get $x=y^2-\frac{3}{2}.$ Accordingly, let $A=\left(a^2-\frac{3}{2},a\right)$ and $B=\left(b^2-\frac{3}{2},b\right).$ We can assume that $a<0$ and $b>0.$



Then the slope of $\overline{AB}$ is $$\begin{array}{rl}\sqrt{3}& =\frac{b-a}{(b^2-\frac{3}{2})-(a^2-\frac{3}{2})}\\ & =\frac{b-a}{b^2-a^2}\\ & =\frac{b-a}{(b-a)(b+a)}\\ & =\frac{1}{a+b}\end{array}$$The difference between the $y$-coordinates of $A$ and $P$ is $a,$ so the difference between the $x$-coordinates of $A$ and $P$ is $\frac{a}{\sqrt{3}}$. Then $$AP=\sqrt{a^2+{\left(\frac{a}{\sqrt{3}}\right)}^2}=\sqrt{\frac{4}{3}{a}^2}=-\frac{2}{\sqrt{3}}a.$$Similarly, $$BP=\frac{2}{\sqrt{3}}b.$$Therefore, $$|AP-BP|=\frac{2}{\sqrt{3}}(a+b)=\frac{2}{\sqrt{3}}\cdot \frac{1}{\sqrt{3}}=\boxed{\frac{2}{3}}.$$