If $f\left(x\right)=\frac{{\left(2^x-1\right)}^3}{\sin \left(\frac{x}{p}\right)\log \left(1+\frac{x^2}{3}\right)}$ is continuous at $x=0$ and $f\left(0\right)={\left(6\log 2\right)}^3$ then the value of  p is

$\underset{x\to 0}{Lt}f\left(x\right)=f\left(0\right)$ 

$$\begin{gathered} \Rightarrow \underset{x\to 0}{Lt}\frac{{(2^x-1)}^3}{\sin \left(\frac{x}{p}\right)\log (1+\frac{x^2}{3})}=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{{\displaystyle \frac{{(2^x-1)}^3}{x^3}}}{{\displaystyle \frac{\sin {\displaystyle \frac{x}{p} }\log (1+{\displaystyle \frac{x^2}{3}})}{x x^2}}} \end{gathered}$$

$\Rightarrow \frac{{\left(\log 2\right)}^3}{\left(\frac{1}{p}\right)\left(\frac{1}{3}\right)}=6{\left(\log 2\right)}^3$

$\Rightarrow 3P=6\Rightarrow P=2$