If all edge weights are $1$-weighted and $K = |V|$ then this problem is the problem of counting hamiltonian cycles in planar graph,which is #P-complete.
But if $K$ is $O(1)$ then let $K = 0$ and all edge weights are $0$. We need to count number of cycles in planar graph (It seems that it is #P-complete also). It is easy to prove that if we can solve this counting problem in polynomail time then we can decide existence of hamiltonian cycle in polynomial time. Let us replace each edge $e=(u, v)$ by two pathes from $u$ to $v$ of length $p = 2|V|^3$, first path is $u,a_1,v_1,a_2,v_2,\ldots,v_{p-1},a_p,v$ and second path is
$u,b_1,v_1,b_2,v_2,\ldots,v_{p-1},b_p,v$, (for different edges $(u,v)$ there are different vertices $a_i,b_i$). Thus for each edge $(u,v)$ we have $2^p$ correponding simple pathes from $u$ to $v$ in new planar graph. There are no hamiltonian path in original graph iff number of cycles in new graph is less then $|V|!\cdot 2^{|V|} \cdot (2^p)^{|V|-1} + |V|^2 \cdot 2^p $ (because at least $(2^p)^{|V|}$ cycles correspond to hamiltinian path in new graph).