Give ellipse

$\frac{x^2}{5}+\frac{y^2}{4}=1 ...\left(1\right)$

Let $P\left(\sqrt{5}\cos \theta , 2\sin \theta \right)$

Ellipse (1)

Given $Q\left(0, -4\right)$

$P{Q}^2={\left(\sqrt{{\left(\sqrt{5}\cos \theta \right)}^2+{\left(2\sin \theta \right)}^2}\right)}^2$

$\Rightarrow P{Q}^2=5{\cos }^2\theta +4{\sin }^2\theta +16{\cos }^2\theta +16$

$\Rightarrow P{Q}^2=4+{\cos }^2\theta +16 \sin \theta +16$

$=1-{\sin }^2\theta +16 \sin \theta$

$\Rightarrow P{Q}^2=17-{\sin }^2\theta +16 \sin \theta$

$\Rightarrow P{Q}^2=17-\left({\sin }^2\theta -16 \sin \theta \right)$

$\Rightarrow P{Q}^2=21\left({\left(\sin \theta -8\right)}^2-64\right)$

$\Rightarrow P{Q}^2=85-(\sin \theta -8{)}^2=85-49$

$\mathrm{Max} P{Q}^2=36$       ($\because$ maximum value of $\sin \theta =1$)