$ \begin{aligned} & \overrightarrow{\mathrm{u}}=((\mathrm{i}+\mathrm{j}) \cdot P Q) P Q \\ & \overrightarrow{\mathrm{u}}=\mid(\mathrm{i}+\mathrm{j}) \cdot P Q \\ & |\overrightarrow{\mathrm{u}}|=\left|(\mathrm{i}+\mathrm{j}) \cdot \frac{(\mathrm{ai}+\mathrm{bj})}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\right|=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}^2+\mathrm{b}^2} \\ & \overrightarrow{\mathrm{v}}=(\mathrm{i}+\mathrm{j}) \cdot P S \\ & |\overrightarrow{\mathrm{v}}|=\left|\frac{(\mathrm{i}+\mathrm{j}) \cdot(\mathrm{ai}-\mathrm{bj})}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\right|=\frac{\mathrm{a}-\mathrm{b}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}} \\ & |\overrightarrow{\mathrm{u}}|=|\overrightarrow{\mathrm{v}}|=|\overrightarrow{\mathrm{w}}| \\ & \frac{|(\mathrm{a}+\mathrm{b})|+|(\mathrm{a}-\mathrm{b})|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}=\sqrt{2} \end{aligned} $
For $\mathrm{a}>\mathrm{b}$
$ \begin{align*} & 2 a=\sqrt{2} \cdot \sqrt{a^2+b^2} \\ & 4 a^2=2 a^2+2 b^2 \\ & a^2=b^2 \therefore a=b \tag{1} \end{align*} $
$ (a>0, b>0) $
similarly for $\mathrm{a}>\mathrm{b}$ we will get $\mathrm{a}=\mathrm{b}$
Now area of parallelogram $=|(\mathrm{ai}+\mathrm{bj}) \times(\mathrm{ai}-\mathrm{bj})|$

$ \begin{align*} & =2 a b \\ & \therefore 2 a b=8 \\ & a b=4 \tag{2} \end{align*} $ from (1) and (2)
$ \mathrm{a}=2, \mathrm{~b}=2 \therefore \mathrm{a}+\mathrm{b}=4 \quad \text { option }(\mathrm{A}) $
length of diagonal is
$ |2 \mathrm{a} \hat{\mathrm{i}}|=|4 \hat{\mathrm{i}}|=4 $
so option (C)