Correct : a

Since P = LU, the system PX = Q becomes LUX = Q. Setting UX = Y, this splits into two simpler systems - first solve LY = Q using forward substitution (L is lower triangular), then solve UX = Y using back substitution (U is upper triangular). Both steps are O(n²) and completely standard. Option A is correct.
Option B - If P is invertible, both L and U are invertible: This is actually also true. Since det(P) = det(L) × det(U), and det(P) ≠ 0 implies both factors are non-zero, both L and U must be invertible. However, the official GATE key marks only Option A.
Option C - If P is singular, at least one diagonal of U is zero: Also true in principle - singular P means det(U) = 0 (since det(L) = 1 for unit lower triangular L), and a zero diagonal in U makes its determinant zero. But again, only A is marked officially correct.
Option D - If P is symmetric, both L and U are symmetric: False. L is lower triangular and U is upper triangular - neither can be symmetric in general. Symmetric matrix decomposition uses Cholesky (P = LL^T), not standard LU.