For the parabola $y^2=4px$, the equation of normal is

$y=mx-2pm-p{m}^3$.

Comparing it with the given equation of normal $y=-\frac{a}{b}x+\frac{1}{b}$, we get

$-\frac{a}{b}=m; -2pm-p{m}^3=\frac{1}{b}$

i.e. $-2p\left(-\frac{a}{b}\right)-p{\left(-\frac{a}{b}\right)}^3=\frac{1}{b}$

$\Rightarrow \frac{2pa}{b}+\frac{p{a}^3}{b^3}=\frac{1}{b}\Rightarrow 2pa{b}^2+p{a}^3=b^2$