If $$t = \sqrt{x} + 4$$, then $$\left(\frac{dx}{dt}\right)_{t=4}$$ is:

We have $$t = \sqrt{x} + 4$$, and we need to find $$\left(\frac{dx}{dt}\right)$$ at $$t = 4$$.

First, we find the value of $$x$$ when $$t = 4$$. Substituting $$t = 4$$ into the equation: $$4 = \sqrt{x} + 4$$, which gives $$\sqrt{x} = 0$$, so $$x = 0$$.

Now we differentiate both sides of $$t = \sqrt{x} + 4$$ with respect to $$t$$: $$1 = \frac{1}{2\sqrt{x}} \cdot \frac{dx}{dt}$$, which gives $$\frac{dx}{dt} = 2\sqrt{x}$$.

At $$t = 4$$, we have $$x = 0$$, so $$\frac{dx}{dt} = 2\sqrt{0} = 0$$.

Hence, the correct answer is Option B.