From the figure
In $\Delta ONA$
Given that angle $NOA=45^{\circ}$
$\Rightarrow\:\cos 45^{\large\circ}=\large\frac{ON}{OA}$
$i.e.,\:\cos 45^{\large\circ}=\large\frac{p}{a}$
$\large\frac{1}{\sqrt 2}=\frac{p}{a}$
$a=\sqrt 2p$
In $\Delta ONB$
angle $NBO=45^{\circ}$
$\Rightarrow\:\cos 45^{\large\circ}=\large\frac{ON}{OB}$
$i.e.,\:\cos 45^{\large\circ}=\large\frac{p}{b}$
$\large\frac{1}{\sqrt 2}=\large\frac{p}{b}$
$b=\sqrt 2 p$
Given that Area of $\Delta OAB=\large\frac{1}{2}$$ab$
$\Rightarrow\:\large\frac{1}{2}$$ab=18$
$\Rightarrow\:\large\frac{1}{2}$$\sqrt 2\sqrt 2p^2=18$
$\Rightarrow\:p^2=18$
or $p=\sqrt{18}$
$\Rightarrow\:p=3\sqrt 2$
Equation of a straight line in normal form is $x\:cos\alpha+y\:sin\alpha=p$
where $\alpha$ is the angle made by the $\perp$ from origin with $x\:axis$
$\therefore$ Equation of the line $AB$ is
$x\:cos 45^{\circ}+y\:sin 45^{\circ}=p$
$i.e.,\:x.\large\frac{1}{\sqrt 2}$$+y.\large\frac{1}{\sqrt 2}$$=3\sqrt 2$
$\Rightarrow\:x+y=6$
Hence (b) is the correct answer.