The term independent of x in the expansion of $\displaystyle \left ( x -\dfrac{1}{4} \right )^4\left ( x + \dfrac{1}{x} \right )^3$

$\begin{aligned} & \left(x-\frac{1}{x}\right)^{4}\left(x+\frac{1}{x}\right)^{3} \\ = & \left(x-\frac{1}{x}\right)\left(x-\frac{1}{x}\right)^{3}\left(x+\frac{1}{x}\right)^{3} \\ = & \left(x-\frac{1}{x}\right)\left\{\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\right\}^{3} \\ = & \left(x-\frac{1}{x}\right)\left(x^{2}-\frac{1}{x^{2}}\right)^{3} \\ = & \left(x-\frac{1}{x}\right)\left(x^{6}-\frac{1}{x^{6}}-3 x^{4} \times \frac{1}{x^{2}}+3 x^{2} \times \frac{1}{x^{4}}\right) \\ = & \left(x-\frac{1}{x}\right)\left(x^{6}-\frac{1}{x^{6}}-3 x^{2}+\frac{3}{x^{2}}\right) \\ = & x^{7}-\frac{1}{x^{5}}-3 x^{3}+\frac{3}{x}-x^{5}+\frac{1}{x^{7}}+3 x-\frac{3}{x^{3}}+0 \end{aligned}$

$\Rightarrow$ Term independent of $x=0$

Hence the correct option is B.