The normal form of the line $x+y+\sqrt{2}=0$  is

$x+y=-\sqrt{2}$

$-x-y=\sqrt{2}$

$\left(\frac{-1}{\sqrt{2}}\right)x+\left(\frac{-1}{\sqrt{2}}\right)y=1$

Compare with $x\cos \alpha +y\sin \alpha =P$

$\cos \alpha =\frac{-1}{\sqrt{2}}$ ,$\sin \alpha =\frac{-1}{\sqrt{2}}$

sin and cos are –ve in 3^rd quadrant.

$\cos \alpha =\cos (180°+45°)=\cos 225°$

$\cos \alpha =\cos \frac{5\pi }{4}$

$\alpha =\frac{5\pi }{4}$

$\therefore$  The normal form of $x+y+\sqrt{2}=0$  is $x\cos \frac{5\pi }{4}+y\sin \frac{5\pi }{4}=1$