Let $$A$$ be a $$2 \times 2$$ matrix with $$\det(A) = -1$$ and $$\det((A + I)(\text{Adj}(A) + I)) = 4$$. Then the sum of the diagonal elements of $$A$$ can be:

We have a $$2 \times 2$$ matrix A with $$\det(A) = -1$$ and $$\det((A + I)(\text{Adj}(A) + I)) = 4$$.

For a $$2 \times 2$$ matrix, $$\text{Adj}(A) = \det(A) \cdot A^{-1} = -A^{-1}$$.

$$(A + I)(\text{Adj}(A) + I) = (A + I)(-A^{-1} + I) = (A + I)(I - A^{-1})$$

$$= A \cdot I - A \cdot A^{-1} + I \cdot I - I \cdot A^{-1}$$

$$= A - I + I - A^{-1} = A - A^{-1}$$

Let $$\text{tr}(A) = s$$. By Cayley-Hamilton: $$A^2 - sA + \det(A) \cdot I = 0$$, so $$A^2 = sA + I$$.

From $$A \cdot A^{-1} = I$$: $$A^{-1} = \frac{1}{\det(A)}(sI - A) = -(sI - A) = A - sI$$.

$$A - A^{-1} = A - (A - sI) = sI$$

$$\det(A - A^{-1}) = \det(sI) = s^2 = 4$$

$$s = \pm 2$$

The sum of diagonal elements of A can be $$2$$ or $$-2$$. From the options, $$2$$ is available.

Therefore, the correct answer is Option B: 2.