Think about the problem visually. One of the ways of understanding a structured linear transformation is studying its action carefully on a generic vector. The following comments are aimed at solving this problem with minimum computation.
The linear transformation $T$ flips the vector vertically along the horizontal axis passing through the 4th co-ordinate.
Clearly if you flip a vector twice, it becomes that vector again.
So $T^2 = I$ and option D is true. And thus option C must be false (Why?).
For option B, observe that for the matrix to be diagonal w.r.t to the standard basis, the vectors of the standard basis must be eigenvectors. Geometrically speaking, a flip cannot scale the length of a vector. Since, $T^2 = I$, eigenvectors of a flip can either fix a vector or scale the vector by $-1$. Therefore a flip cannot maintain the same direction, unless the vectors are symmetric/anti-symmetric about the horizontal axis passing through 4th co-ordinate. Hence the standard vectors cannot be eigenvectors.
However, the above reasoning tells us that there are four linearly independent eigenvectors associated with eigenvalue $1$:
and 3 linearly independent eigenvectors associated eigenvalue $-1$:
Therefore $\det(T) = (-1)^3(1)^4 = -1$ and option A is wrong.