 Solution

Reverse time under the uniform invariant law. The reverse shuffle selects $k$ cards uniformly without replacement and moves them to the top in random order. Track cards once selected. The forward time $\tau$ at which the initially bottom card first lies among the next top $k$ corresponds in reverse to the first time every card has been selected. At that coupon-collector time, the order of all marked cards is uniform and independent of the marking time, by induction over the random insertions. Reversing again shows that $X_{\tau }$ is uniform and independent of $\tau$, so part (a) makes $\tau$ a strong stationary time.