If $$P = \begin{bmatrix}a & b & c\\x & y & z\\p & q & r \end{bmatrix}$$ and  $$Q = \begin{bmatrix}-x & a & -p \\y & -b & q\\z & -c & r \end{bmatrix}$$ then

Taking -1 common from both 2nd and 3rd row of Q, we get

Q = (-1)(-1)$$\begin{bmatrix}-x & a & -p \\-y & b & -q\\-z & c & -r \end{bmatrix}$$

= $$\begin{bmatrix}-x & a & -p \\-y & b & -q\\-z & c & -r \end{bmatrix}$$

Now taking -1 common from 1st and 3rd column of Q, we get

Q= (-1)(-1)($$\begin{bmatrix}x & a & p \\y & b & q\\z & c & r \end{bmatrix}$$

= ($$\begin{bmatrix}x & a & p \\y & b & q\\z & c & r \end{bmatrix}$$

After interchanging the 1st and second column, Q becomes the the transpose of P.

Since the row have been exchange once and the determinant of the transpose matrix is the same, hence det(P)=(-1)$$^1$$*det(Q)

=> $$det(P) = -det(Q)$$

 D is the answer.