$\text{Total number of ordered pairs}(x,y)\text{satisfying}$
$|x|+|y|=2,\sin \left(\frac{\pi x^2}{3}\right)=1\text{is}$

The correct option is C

$4$

Given
$|x|+|y|=2\Rightarrow |x|,|y|\in [0,2]$

$\sin \left(\frac{\pi x^2}{3}\right)=1\Rightarrow \frac{\pi x^2}{3}=\frac{(4n+1)\pi }{2}$
$\Rightarrow \frac{x^2}{3}=\frac{(4n+1)}{2}$
$\Rightarrow x^2=(4n+1)\frac{3}{2}$
$[\because |x|\le 2\Rightarrow x^2\le 4]$
$\Rightarrow x^2=\frac{3}{2}$
$\Rightarrow |x|=\sqrt{\frac{3}{2}}\Rightarrow |y|=2-\frac{\sqrt{3}}{2}$
$\Rightarrow x=+\sqrt{\frac{3}{2}},-\sqrt{\frac{3}{2}}y=(2-\frac{\sqrt{3}}{2}),-(2-\frac{\sqrt{3}}{2})$
$\therefore 4\text{Ordered pairs are possible.}$
Hence the correct answer is Option C.