If $(x, y)$ is old coordinates and $(X, Y)$ are new coordinates, when axes are rotated through an angle of $\theta$, then
$$
\begin{aligned}
& x=X \cos \theta-Y \sin \theta \text { and } y=X \sin \theta+Y \cos \theta \\
& \therefore \quad x=2 \cos 135^{\circ}-(-6) \sin 135^{\circ} \\
& \text { and } \quad y=2 \sin 135^{\circ}+(-6) \cos 135^{\circ} \\
& \Rightarrow x=2\left(\frac{-1}{\sqrt{2}}\right)+6\left(\frac{1}{\sqrt{2}}\right) \text { and } y=2\left(\frac{1}{\sqrt{2}}\right)-6\left(\frac{-1}{\sqrt{2}}\right) \\
& \Rightarrow x=\frac{4}{\sqrt{2}} \text { and } y=\frac{8}{\sqrt{2}} \\
& \Rightarrow x=2 \sqrt{2} \text { and } y=4 \sqrt{2} .
\end{aligned}
$$
$\therefore$ Coordinates of $P$ in the original system are $(2 \sqrt{2}, 4 \sqrt{2})$.