Step 1: Recall the normal at a parameter.
For $y^2=4px$, the normal at parameter $t$ is $y=-tx+2pt+pt^3$.
Step 2: Put the given line in slope form.
The line $ax+by=1$ becomes $y=-\dfrac{a}{b}x+\dfrac{1}{b}$.
Step 3: Match the slopes.
Comparing slopes, $-t=-\dfrac{a}{b}$, so $t=\dfrac{a}{b}$.
Step 4: Match the intercepts.
Comparing constant terms, $\dfrac{1}{b}=2pt+pt^3$. Substituting $t=\dfrac{a}{b}$ gives $\dfrac{1}{b}=\dfrac{2pa}{b}+\dfrac{pa^3}{b^3}$.
Step 5: Clear denominators.
Multiplying by $b^3$, $b^2=2pab^2+pa^3$.
Step 6: Isolate the condition.
Rearranging, $pa^3=b^2-2pab^2$.
\[ \boxed{pa^3=b^2-2pab^2} \]