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