The co-ordinates of $P$ are $\left(1,0\right).$

We know that, any point on the parabola $y^2=4ax$ is $\left(a{t}^2,2at\right).$

 Hence, for the parabola $y^2=8x$ the point $Q$ is $\left(2t^2,4t\right).$

Let mid-point of $PQ$ be $\left(h,k\right),$ then by using mid-point formula, we get $h=\frac{2t^2+1}{2}$

 $\Rightarrow 2h=2t^2+1 \dots \left(1\right)$

and $k=\frac{4t+0}{2}$ $\Rightarrow t=\frac{k}{2} \dots \left(2\right)$

On putting the value of $t$ from Equation $\left(2\right)$ in Equation $\left(1\right),$ we get

$2h=2{\left(\frac{k}{2}\right)}^2+1,$ $\Rightarrow 2h=\frac{2k^2}{4}+1,$

$\Rightarrow 4h=k^2+2$

The locus is obtained by replacing $\left(h,k\right)$ by $\left(x,y\right).$

Hence, the locus is $y^2-4x+2=0.$