If the line y - 2 = 0 is the directrix of the parabola x ^ 2 - ky +...

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