Step-by-step Solution:

1. Recall the property of adjoint matrices: For any $n \times n$ matrix $A$, the determinant of its adjoint is related to the determinant of the matrix itself by the formula: $\det(\text{adj}(A)) = (\det(A))^{n-1}$

2. Apply the property to the given matrix P: In this problem, P is a $3 \times 3$ matrix, so $n=3$. The formula becomes: $\det(\text{adj}(P)) = (\det(P))^{3-1} = (\det(P))^2$

3. Calculate the determinant of the given adjoint matrix: We are given the adjoint of P as: $\text{adj}(P) = \left[ {\begin{matrix} 1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3 \\ \end{matrix} } \right]$ Let's calculate its determinant: $\det(\text{adj}(P)) = 1 \begin{vmatrix} 1 & 7 \\ 1 & 3 \end{vmatrix} - 4 \begin{vmatrix} 2 & 7 \\ 1 & 3 \end{vmatrix} + 4 \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix}$ $\det(\text{adj}(P)) = 1((1)(3) - (7)(1)) - 4((2)(3) - (7)(1)) + 4((2)(1) - (1)(1))$ $\det(\text{adj}(P)) = 1(3 - 7) - 4(6 - 7) + 4(2 - 1)$ $\det(\text{adj}(P)) = 1(-4) - 4(-1) + 4(1)$ $\det(\text{adj}(P)) = -4 + 4 + 4$ $\det(\text{adj}(P)) = 4$

4. Solve for the determinant of P: Substitute the value of $\det(\text{adj}(P))$ into the equation from Step 2: $(\det(P))^2 = 4$ Taking the square root of both sides, we get: $\det(P) = \pm \sqrt{4}$ $\det(P) = \pm 2$

5. Conclusion: The possible values for the determinant of P are 2 and -2.

Evaluating the Options:

• A: -2 - This is a possible value.
• B: -1 - This is not a possible value.
• C: 1 - This is not a possible value.
• D: 2 - This is a possible value.

Therefore, the correct options are A and D.