Given the equation of an ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$,where $a=2, b=\sqrt{3}.$
As we know equation of normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ can be taken as
$ax\sec \theta -by\mathrm{cosec}\theta =a^2-b^2$
$\Rightarrow 2x\sec \theta -\sqrt{3}y\mathrm{cosec}\theta =1$ for the given ellipse.
Now, this normal is parallel to $2x+y=5$.
$\Rightarrow -\frac{2\sec \theta }{\sqrt{3}\mathrm{cosec}\theta }=-2\Rightarrow \tan \theta =-\sqrt{3}\Rightarrow \theta =\frac{2\pi }{3},-\frac{\pi }{3}$
Hence point $P\left(a\cos \theta , b\sin \theta \right)=\left(-1,\frac{3}{2}\right)$ or $\left(1, -\frac{3}{2}\right)$
Tangent at $\left(-1,\frac{3}{2}\right)$ is $\frac{x.\left(-1\right)}{4}+\frac{y.\frac{3}{2}}{3}=1$
$\Rightarrow -x+2y=4$ which also passes through $Q\left(4, 4\right)$
$\Rightarrow PQ=\sqrt{{\left(4\pm 1\right)}^2+{\left(4\mp \frac{3}{2}\right)}^2}=\frac{5\sqrt{5}}{2}.$