In a switch-tail ring counter with a single flip flop, $\bar Q$ is connected back as the input. Since, the FF is $T,$ $Q = \bar D$ and $\bar Q = D.$
Now, if the initial value of $Q = 1, \bar Q = 0.$ In next clock, $T = 0.$ So, no toggle and $Q = 1, \bar Q = 0.$
if the initial value of $Q = 0, \bar Q = 1.$ In next clock, $T = 1.$ So, toggle and $Q = 1, \bar Q = 0.$
So, within $2$ cycles output becomes $1$ and does not change afterwards. So, correct option is D – None of the above.