Let $P$ be a point on parabola $y^2=8 x$ whose coordinates are $\left(2 t^2, 4 t\right)$ and $A$ is the point $(1,0)$.
Let $(x, y)$ be the mid-point of $P\left(2 t^2, 4 t\right)$ and $A(1,0)$.
$$
\begin{aligned}
& \therefore & x=\frac{2 t^2+1}{2} \\
\text { and } & y & =\frac{4 t+0}{2}=2 t
\end{aligned}
$$
On eliminating $t$, we get
$$
\begin{array}{rlrl}
& & x & =\frac{y^2+2}{4} \\
\Rightarrow & 4 x & =y^2+2 \\
& y^2 & =4\left(x-\frac{1}{2}\right)
\end{array}
$$
Hence, the locus of the mid-point of the line segment $A P$ is $y^2=4\left(x-\frac{1}{2}\right)$.