If the line $\mathrm{y}-2=0\mathrm{}$ is the directrix of the parabola ${\mathrm{x}}^2-\mathrm{k}\mathrm{y}+32=0,\mathrm{k}\ne 0\mathrm{}$ and the parabola intersects the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=8\mathrm{}$ at two real distinct points, then the absolute value of $\mathrm{k}$ is

${\mathrm{x}}^2-\mathrm{k}\mathrm{y}+32=0$
$\Rightarrow {\mathrm{x}}^2=\mathrm{k}\left(\mathrm{y}-\frac{32}{\mathrm{k}}\right)$
Put, $\mathrm{x}=\mathrm{X},\mathrm{y}-\frac{32}{\mathrm{k}}=\mathrm{Y}$
The equation of directrix is $\mathrm{Y}+\frac{\mathrm{k}}{4}=0$
i.e. $\mathrm{y}-\frac{32}{\mathrm{k}}+\frac{\mathrm{k}}{4}=0$
But, $\mathrm{y}-2=0\mathrm{}$ is the directrix.
$\Rightarrow \frac{32}{\mathrm{k}}-\frac{\mathrm{k}}{4}=2$
$\Rightarrow {\mathrm{k}}^2+8\mathrm{k}-128=0$
$\Rightarrow \mathrm{k}=-16$ or $\mathrm{k}=8$
For $\mathrm{k}=8,$ the parabola is ${\mathrm{x}}^2=8\left(\mathrm{y}-4\right)$ which does not intersect the circle.
For $\mathrm{k}=-16,$ the parabola is ${\mathrm{x}}^2=-16\left(\mathrm{y}+2\right)$ which intersects the circle at two real distinct points.
$\mathrm{}\Rightarrow \mathrm{}$ Absolute value of $\mathrm{k}=\left|-16\right|=16$