The equation of the tangent at $P \left(2,8\right)$ to the parabola is

$\frac{y+8}{2}=2x-\left(x+2\right)+8$
$\Rightarrow 2x-y+4=0 ...\left(\mathrm{i}\right)$ 

So, the slope of the normal to the circle is $-\frac{1}{2}$
The equation of normal to the circle at $Q$ will be

$y+7=-\frac{1}{2}\left(x+9\right)$
$x+2y+23=0 ...\left(\mathrm{ii}\right)$
The point $Q$ is the intersection of the tangent at $P$ and the normal at $Q$

On solving the equation $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$, we get,

$x+2\left(2x+4\right)+23=0$

$\Rightarrow 5x=-31\Rightarrow x=-\frac{31}{5}$

and $y=2\left(-\frac{31}{5}\right)+4=-\frac{42}{5}$

Hence, the coordinates of the point $Q$ are $\left(-\frac{31}{5},-\frac{42}{5}\right)$