Closed Walk in Planar Graphs
It's not, since there can be exponentially many such walks.

The $((2 \cdot n)+1)$ by $((2 \cdot n)+1)$ grid graph will have at least $2^n$  such paths for $​k = (2\cdot n \cdot n)+(4 \cdot n)+1$, since paths can go along two sides of the grid, in from both ends of that, then connect both ends of that in a way which encloses exactly two squares of each of the $n$ non-overlapping 2-by-2 blocks on the diagonal between those ends.

However, there are at least 8 weaker versions of what you're asking about.


Enumerating them in poly(input_size, number_of_such_paths) time.
Enumerating them in ​poly(input_size) $\cdot$ number_of_such_paths time.
Enumerating them with polynomial delay.
Doing one of the previous three while using only poly(input_size) space.
Computing number_of_such_paths and a bijection from {0,1,2,3,...,number_of_such_paths-1} to the set of such paths.
Doing the previous thing and computing the inverse of the bijection.


I don't know anything about whether or not any of those are doable in polynomial time.