To determine $$g^{\prime}(0)$$, we start by applying the chain rule and product rule to find the derivative of the given function $$g(x) = h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$$.

The product rule states that if we have two functions $$u(x)$$ and $$v(x)$$, then the derivative of their product is given by:

$$ (u(x)v(x))' = u'(x)v(x) + u(x)v'(x) $$

Let's denote $$u(x) = h(\mathrm{e}^x)$$ and $$v(x) = \mathrm{e}^{h(x)}$$.

First, we need to find $$u'(x)$$ and $$v'(x)$$. Using the chain rule, we find:

$$ u'(x) = \frac{d}{dx}h(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \frac{d}{dx}(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \mathrm{e}^x $$

Now, the derivative of $$v(x)$$ is:

$$ v'(x) = \frac{d}{dx}(\mathrm{e}^{h(x)}) = \mathrm{e}^{h(x)} \cdot h'(x) $$

Using the product rule, we get the derivative of $$g(x)$$:

$$ g'(x) = u'(x) v(x) + u(x) v'(x) $$

Substituting $$u(x)$$, $$v(x)$$, $$u'(x)$$, and $$v'(x)$$ into the above expression, we get:

$$ g'(x) = \left( h'(\mathrm{e}^x) \mathrm{e}^x \right) \mathrm{e}^{h(x)} + \left( h(\mathrm{e}^x) \right) \left( \mathrm{e}^{h(x)} h'(x) \right) $$

Next, we need to evaluate this at $$x = 0$$:

First, we know that:

$$ h(0) = 0 $$

$$ h(1) = 1 $$

$$ h'(0) = 2 $$

$$ h'(1) = 2 $$

Substituting $$x = 0$$ into the expressions, we get:

$$ u(0) = h(\mathrm{e}^0) = h(1) = 1 $$

$$ v(0) = \mathrm{e}^{h(0)} = \mathrm{e}^0 = 1 $$

$$ u'(0) = h'(\mathrm{e}^0) \mathrm{e}^0 = h'(1) \cdot 1 = 2 $$

$$ v'(0) = \mathrm{e}^{h(0)} h'(0) = \mathrm{e}^0 \cdot 2 = 2 $$

Therefore, evaluating $$g'(0)$$:

$$ g'(0) = \left( u'(0) v(0) \right) + \left( u(0) v'(0) \right) $$

$$ g'(0) = \left( 2 \cdot 1 \right) + \left( 1 \cdot 2 \right) $$

$$ g'(0) = 2 + 2 = 4 $$

Thus, the value of $$g^{\prime}(0)$$ is 4, which corresponds to Option A.

The correct answer is Option A: 4.