There are $N$ accesses to cache.
Out of these $n$ are unique block addresses.
Now, we need to find the number of misses. (min. $n$ misses are guaranteed whatever be the access sequence due to $n$ unique block addresses).
We are given that between two consecutive accesses to the same block, there can be only $k$ unique block addresses. So, for a block to get replaced we can assume that all the next $k$ block addresses goes to the same set (given cache is set-associative) which will be the worst case scenario (they may also go to a different set but then there is lesser chance of a replacement). Now, if associativity size is $\geq k$, and if we use LRU (Least Recently Used) replacement policy, we can guarantee that these $k$ accesses won't throw out our previously accessed cache entry (for that we need at least k accesses). So, this means we are at the best-cache scenario for cache replacement -- out of $N$ accesses we miss only $n$ (which are unique and can not be helped from getting missed and there is no block replacement in cache). So, miss ratio is $n/N$.
PS: In question it is given "bounded above by $k$", which should mean $k$ unique block accesses as $k$ is an integer, but to ensure no replacement this must be '$k-1$'. Guess, a mistake in question.
Correct Answer: $A$