Given, equation of circle is
$x^{2}+y^{2}+2 x-2 y-2=0$
$\Rightarrow (x+1)^{2}+(y-1)^{2}=4 $
$\therefore $ Centre $(-1,1)$ and radius $=2$
Let $(h, k)$ be the mid-point of chord.
From figure,
$O P=\sqrt{(h+1)^{2}+(k-1)^{2}}$
In $\triangle O A P$,
$\sin 45^{\circ}=\frac{O P}{O A}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{\sqrt{(h+1)^{2}+(k-1)^{2}}}{2}$
On squaring both sides, we get
$ (h+1)^{2}+(k-1)^{2}=2 $
$\Rightarrow h^{2}+k^{2}+2 h-2 k=0$
$\therefore $ Locus of $P$ will be
$x^{2}+y^{2}+2 x-2 y=0$
