Given, parabola, $y^{2}=8x$
$\Rightarrow 2 y \frac{d y}{d x}=8$
$\Rightarrow \frac{d y}{d x}=\frac{4}{y}$
Slope of normal of parabola $=-\frac{y}{4}\, \dots(i)$
Given, $2x+y+\lambda=0$ is a equation of normal of the parabola $y^{2}=8x$
$\therefore $ Slope of normal $=-2\, \dots(ii)$
From Eqs. (i) and (ii), we get
$-\frac{y}{4}=-2$
$ \Rightarrow y=8$
$\therefore (8)^{2}=8 x$
$ \Rightarrow x=8$
Now, putting the values of x and y in the equation of normal
$2 (8) +8+ \lambda =0$
$\Rightarrow \lambda=-24$