For option $(A):$ A matrix $A_{n \times n}$ is called skew-symmetric if $A^{T} = -A.$ Here $A^{T}$ is the transpose of $A.$
We know that $\mid A \mid = \mid A^{T} \mid $
$\implies \mid A \mid = \mid -A \mid $
$\implies \mid A \mid = (-1)^{n} \mid A \mid$
If $n$ is odd, then $\mid A \mid = - \mid A \mid \implies \mid A \mid = 0$
$\therefore$ Every odd degree skew-symmetric matrix is not invertible, or equivalently singular.
If $n$ is even, the determinant of a skew-symmetric matrix can always be written as the square of a polynomial in the matrix entries, a polynomial with integer coefficients that only depend on the size of the matrix.
Ref:
https://en.wikipedia.org/wiki/Pfaffian
For even order, let's $A = \begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix} \implies \mid A \mid = 0 - (-a\cdot a) = a^{2}$
So, the determinant of a skew-symmetric matrix over real numbers is non-negative.
For option $(B):$ The diagonal of skew-symmetric matrix consists of zero elements and therefore the sum of elements in the main diagonals is equal to zero. When identity matrix is added to skew-symmetric matrix then all the diagonal elements become $1$ and the resultant matrix is invertible.
Let $A = \begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix} \implies \mid A \mid = 0 - (-a\cdot a) = a^{2} \neq 0$
Now, $A + I = \begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\implies A + I = \begin{bmatrix} 1 & a \\ -a & 1 \end{bmatrix} \implies \mid A + I \mid = 1 + a^{2} \neq 0$
So, the matrix $A + I$ is invertible.
For option $(C),$ a counter example is enough.
$A = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}_{2 \times 2}, B = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}_{2 \times 2}$
Now, $AB = \begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}_{2 \times 2}$
For option $(D),$ a counter example is enough.
$A = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}_{2 \times 2}, B = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}_{2 \times 2}$
Now, $AB = \begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}_{2 \times 2}$
and, $BA = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}_{2 \times 2}$
So, the correct answer is $A;B.$