$\begin{array}{ll}
\because & A \text { and } B \text { lies on } x+y=1 \\
\therefore & A(1,0), B(0,1)
\end{array}$
also $O P \perp A B$
$\begin{aligned}
& \therefore \quad m_{O P} \times m_{A B}=-1 \\
& \Rightarrow \quad \frac{k-0}{h-0} \times-1=-1 \\
& \therefore \quad \frac{k}{h}=1 \Rightarrow k=h
\end{aligned}$
also $D$ is mid point of $O P$.
$\therefore \quad D\left(\frac{h+0}{2}, \frac{k+0}{2}\right) \equiv D\left(\frac{h}{2}, \frac{k}{2}\right)$
$\because \quad D$ lies on $A B$
$\begin{array}{ll}
\therefore & \frac{h}{2}+\frac{k}{2}=1 \\
\Rightarrow & h+k=2 \\
\because & h=k \Rightarrow h=k=1 .
\end{array}$
$\therefore \quad$ The image will be circle with centre $(h, k)$ i.e. $(1,1)$ and radius same as that of $x^2+y^2=1$
$\therefore \quad$ Locus of image :
$\begin{aligned}
& (x-1)^2+(y-1)^2=1 \\
\Rightarrow & x^2+y^2-2 x-2 y+1=0
\end{aligned}$