The area of the region enclosed by the curve $f\left(x\right)=\max \{\sin x,\cos x\},-\pi \le x\le \pi$ and the $x-$axis is

Plotting the required diagram we get,

Now from diagram we can say that,

$\sin x$ is maximum for $x\in \left(-\mathrm{\pi },\frac{-3\mathrm{\pi }}{4}\right)\cup \left(\frac{\mathrm{\pi }}{4},\mathrm{\pi }\right)$ and $\cos x$ is maximum for $x\in \left(\frac{-3\mathrm{\pi }}{4},\frac{-\mathrm{\pi }}{2}\right)\cup \left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{4}\right)$

So, area of the region is given by,

$A=\left|{\int }_{-\pi }^{\frac{-3\pi }{4}}\sin xdx\right|+\left|{\int }_{\frac{\pi }{4}}^{\pi }\sin xdx\right|+\left|{\int }_{\frac{-3\pi }{4}}^{\frac{-\pi }{2}}\cos xdx\right|+\left|{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{4}}\cos xdx\right|$

$\Rightarrow A=\left|\frac{1}{\sqrt{2}}-1\right|+\left|1+\frac{1}{\sqrt{2}}\right|+\left|-1+\frac{1}{\sqrt{2}}\right|+\left|\frac{1}{\sqrt{2}}+1\right|$

$\Rightarrow A=2+\frac{2}{\sqrt{2}}+2-\frac{2}{\sqrt{2}}=4$

Hence this is the required option.