To detect the fault, we should get an unexpected output. The final gate here is a NOR gate which produces output 0 if either of its input is 1 and else 1. i.e., the output will be 0 for inputs $(0,1), (1,0)$ and $(1,1)$ and output will be 1 for (0,0).
By grounding $T$ is at 0. So, we can ignore the inputs $(1,0)$ and $(0,0)$ to the final NOR gate as they won't be detecting faults. Now, expected $(1,1)$ input will become $(1,0)$ due to grounding of $T$ but produces same output $0$ as for $(1,1)$. Hence this also cannot detect the defect. So, to detect the defect, the input to the final gate must be $(0,1)$ which is expected to produce a 0 but will produce a 1 due to grounding of $T$.
Now, for $(0,1)$ input for the final gate, we must have,
$X_3 = X_4 = 1$
But if $X_4=1,$the OR gate makes 1 output and we won't get $(0,1)$ input for the final gate. This means, no input sequence can detect the fault of the circuit.
Alternatively, we can write equation for the circuit as
$(((x_1.x_2.x_3)' + x_4) + x_3.x_4)' \\= ((x_1.x_2.x_3)' + x_4)' . (x_3.x_4)' \\=x_1.x_2.x_3 . x_4' . (x_3'+x_4') \\= x_1.x_2.x_3 . x_4'$
For the faulty circuit output will be
$((x_1.x_2.x_3)' + x_4)' = x_1.x_2.x_3 . x_4'$
So, there is no effect of $T$ being grounded here. Answer is D option.