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

in ring counter(modn), only one bit is high(1) at every count out of n bits and here also due to decoder, only one bit is high at every count out of k bits.

and we can sum up like this -> "the above circuit is counting k distinct states and at each state only one bit is high out of k bits".......this is property of ring counter

Binary counter of $n$ bits can count up to $2^n$ numbers. When this output from counter is fed as input (n bit) to decoder one out of $2^n$ output lines will be activated. So, this arrangement of counter and decoder is behaving as $2^n$ or $k\text{-bit}$ ring counter.

@Shiva Chaitanya Gaju If there are 2 bits then ring counter will be able to count only 2 states right how can it count 2power n states to count 2 power n states it should be a ripple counter right

Initial input n is to decoder but the option says k bit counter and k = 2^n . Decoder will give 2^n unique outputs and a 2^n ring counter too will give 2^n counts, hence the answer.

A circuit is built by giving the output of an $n-bit$ binary counter as input to an $n-to-2^n$ bit decoder.

Output of $n-bit$ binary counter gives $2^n$ output. Number of variable in this $2^n$ outputs is $n$. Decoder generates minterms of the input, for $n$ input there are $2^n$ minterms $(0, 1, ..., 2^{n-1})$ which are nothing but states. Now, since output of binary counter generates $2^n$ outputs in sequence, the outputs of decoder are also in sequence. So, these $2^{n}$ states are minterms & are in sequence which is the output of $2^n$ bit ring counter.

For output of a decoder , only single output will be ‘1’ and remaining will be ‘0’ at the same time. So high output will give the count of the ring counter. Hence Ans is ( C) part.