Equation of tangent at $\left(2, -4\right) \left(T=0\right)$

$-4y=4\left(x+2\right)$

$x+y+2=0 ...\left(1\right)$

Equation of normal

$x-y+\lambda =0$

$\left(2, -4\right)$

$\lambda =-6$

Thus,

$x-y=6 ...\left(2\right)$

Equation of normal

$\mathrm{POI}$ of $\left(1\right) \& x=-2$ is $A\left(-2, 0\right)$

$\mathrm{POI}$ of $\left(2\right) \& x=-2$ is $B\left(-2, -8\right)$

Given 

$AQBP$ is a square

$m_{AQ}·m_{AP}=-1$

$\frac{b}{a+2}\times \frac{4}{-4}=-1$

$a+2=b$

Also $PQ$ must be parallel to $x-$axis

thus,

$b=-4$

$a=-6$

Thus, $2a+b=-16$