Let $k = 2^n$. A circuit is built by giving the output of an $n$-bit binary counter as input to an $n$-to-$2^n$ bit decoder. This circuit is equivalent to a

An $n$-bit binary counter has $2^n = k$ distinct states $(0, 1, \dots, k-1)$.

An $n$-to-$2^n$ decoder takes $n$ inputs and activates exactly one of its $2^n = k$ output lines corresponding to the binary value of the input.

As the $n$-bit binary counter cycles through all its $k$ counts, the decoder activates output line $0$, then line $1$, $\dots$, up to line $k-1$, and then wraps around to line $0$. At any point in time, exactly one of the $k$ lines is active (HIGH), and this active state rotates sequentially across all $k$ outputs.

This behavior matches a $k$-bit ring counter, which consists of $k$ flip-flops with a single circulating active bit (1).

Therefore, (C) is the correct option.