Answer: C
There can be many interpretations of rank of a matrix. Some of which are:
Number of linearly independent vectors in the matrix.
Number of pivots in row echelon form of the matrix.
What does $det(A) = 0$ mean?
Determinant of a matrix is zero if and only if it is not a full rank matrix, i.e. if $A$ is an $n \times n$ matrix and $det(A) = 0$ then the rank of the matrix has be $< n$. This eliminates the option D.
Now comes the tricky part to filter out other options using the information given. $Adj(A)$ is not $null$
$Adj(A)$ is the transpose of the cofactor matrix of $A$. What did I just say?
Co-factor matrix is a matrix having the co-factors as the elements of the matrix.
Co-factor of an element within the matrix is obtained when the minor $M_{ij}$ of the element is multiplied with $(-1)i+j$. Here $i$ and $j$ are the positional values of the element and refers to the row and the column to which the given element belongs. The co-factor of the element is denoted as $C_{ij}$. If the minor of the element is $M_{ij}$, then the co-factor of element would be: $C_{ij} = (-1)^{i+j}) M_{ij}$.
Let's understand this with an example.
$A = \left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{array}\right]$
Then the minor of the element $a_{12}$ is: $M_{12} = \begin{vmatrix}
a_{21} & a_{23}\\
a_{31} & a_{33}
\end{vmatrix}$.
Doing this for all the elements of matrix $A$ we get:
$\begin{align}\text{Co-factor Matrix}&=\left[\begin{array}{ccc} (-1)^{1 + 1}M_{11} & (-1)^{1 + 2}M_{12} & (-1)^{1 + 3}M_{13} \\ (-1)^{2 + 1}M_{21} & (-1)^{2 + 2}M_{22} & (-1)^{2 + 3}M_{23} \\ (-1)^{3 + 1}M_{31} & (-1)^{3 + 2}M_{32} & (-1)^{3 + 3}M_{33} \end{array}\right] \\&=\left[\begin{array}{ccc} +M_{11} & -M_{12} & +M_{13} \\ -M_{21} & +M_{22} & -M_{23} \\ +M_{31} & -M_{32} & +M_{33} \end{array}\right] \\& = \left[\begin{array}{ccc} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{array}\right] \end{align}$.
Transposing this co-factor matrix i.e. rows are exchanged as columns we get the $Adj(A)$.
What does it mean by having the $Adj(A)$ not null?
For starters this means that some of the elements in the $Adj(A)$ are non zero. (For this question it doesn’t really matter if we talk w.r.t. co-factor matrix or the $Adj(A)$, because determinant is same for $A$ and $A^{T}$).
Looking back at the formula where we considered $3 \times 3$ matrix, we see that co-factor of an element is nothing but the determinant of corresponding $2 \times 2$ matrix. (of course it may be negative or positive depending upon the values of $i$ and $j$ but this detail is not required) which brings us to the square one.
What they are saying with this statement is that there exists a $3 \times 3$ matrix inside the given $4 \times 4$ matrix whose determinant is non zero. Meaning the rank of that $3 \times 3$ sub-matrix is $3$. So the rank of the original matrix $A$ has to be $\geq 3$, why? That I'll leave for you to figure out, here is a hint: https://math.stackexchange.com/a/3832359
But we already concluded that it has to be $< 4$. So the rank of the given matrix is $3$.
