If $$\displaystyle\lim_{x \to 0} \left(\dfrac{\tan x}{x}\right)^{1/x^2} = p$$, then $$96 \log_e p$$ is equal to ________.

We need to find $$p = \displaystyle\lim_{x \to 0} \left(\dfrac{\tan x}{x}\right)^{1/x^2}$$.

Taking the natural logarithm: $$\ln p = \displaystyle\lim_{x \to 0} \dfrac{1}{x^2} \ln\left(\dfrac{\tan x}{x}\right)$$.

Using the Taylor series: $$\tan x = x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + \ldots$$

$$\dfrac{\tan x}{x} = 1 + \dfrac{x^2}{3} + \dfrac{2x^4}{15} + \ldots$$

$$\ln\left(\dfrac{\tan x}{x}\right) = \ln\left(1 + \dfrac{x^2}{3} + \dfrac{2x^4}{15} + \ldots\right)$$

Using $$\ln(1 + u) = u - \dfrac{u^2}{2} + \ldots$$ where $$u = \dfrac{x^2}{3} + \dfrac{2x^4}{15} + \ldots$$:

$$\ln\left(\dfrac{\tan x}{x}\right) = \dfrac{x^2}{3} + \dfrac{2x^4}{15} - \dfrac{1}{2}\left(\dfrac{x^2}{3}\right)^2 + \ldots = \dfrac{x^2}{3} + \dfrac{2x^4}{15} - \dfrac{x^4}{18} + \ldots$$

Therefore: $$\ln p = \displaystyle\lim_{x \to 0} \dfrac{1}{x^2}\left(\dfrac{x^2}{3} + O(x^4)\right) = \dfrac{1}{3}$$

So $$p = e^{1/3}$$.

$$96 \log_e p = 96 \cdot \dfrac{1}{3} = 32$$.

Hence, the correct answer is 32.