$\{x \in \mathbb{R} \mid \frac{\sqrt{|x|^2 - 2|x| - 8}}{\log(2 - x - x^2)}\}$ is a real number.

Let

$f\left(x\right)=\frac{\sqrt{{\left|x\right|}^2-2\left|x\right|-8}}{\log \left(2-x-x^2\right)}$

$\Rightarrow f\left(x\right)=\frac{\sqrt{{\left|x\right|}^2-2\left|x\right|-8}}{\log \left(-x^2-x+1\right)}$

$\log \left(-x^2-x+1\right)$ is defined for 

$-x^2-x+1>0$

$\Rightarrow x^2+x-1<0$

$\Rightarrow \left(x-\left(\frac{-1+\sqrt{5}}{2}\right)\right)\left(x-\left(\frac{-1-\sqrt{5}}{2}\right)\right)<0$

$\Rightarrow x\in \left(\frac{-1-\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2}\right)$

$\Rightarrow x\in \left(-1.61,0.61\right) ....\left(\mathrm{i}\right)$

And,

${\left|x\right|}^2-2\left|x\right|-8\ge 0$

Put $\left|x\right|=t$

$\Rightarrow t^2-2t-8\ge 0$

$\Rightarrow \left(t+2\right)\left(t-4\right)\ge 0$

$\Rightarrow t\in (-\infty ,-2]\cup [4,\infty )$

$\Rightarrow \left|x\right|\in (-\infty ,-2]\cup [4,\infty )$

But $\left|x\right|\ge 0 \forall x\in R$, so

$\Rightarrow \left|x\right|\in [4,\infty )$

$\Rightarrow x\in (-\infty ,-4]\cup [4,\infty ) ....\left(\mathrm{ii}\right)$

But $\left(\mathrm{i}\right) \cap \left(\mathrm{ii}\right)$ is $\varphi$, so option $\left(B\right)$ is correct.