$P(-3,2)$ is an end point of focal chord
$P Q$ of the parabola $y^2+4 x+4 y=0$
$(y+2)^2=-4(x-1)$
Let $x-1=-t^2 y+2=-2 t \Rightarrow y=-2 t-2$
$\begin{aligned}
& \Rightarrow & x & =1-t^2, y=-2 t-2 \\
& \therefore & -3 & =1-t^2, 2=-2 t-2 \Rightarrow t= \pm 2, t=-2 \\
& \therefore & t & =-2
\end{aligned}$
End point of focal chord
$\begin{aligned}
& \therefore \text { point } Q\left(\frac{3}{4},-3\right) \\
& y^2+4 x+4 y=0 \Rightarrow 2 y \frac{d y}{d x}+4+4 \frac{d y}{d x}=0 \\
& \frac{d y}{d x}=\frac{-4}{2 y+4}=\left(\frac{d y}{d x}\right)_{\left(\frac{3}{4},-3\right)}=\frac{-4}{-2}=2
\end{aligned}$
$\therefore$ Slope of normal of $\theta=\frac{-1}{2}$