Permutation

Consider the cycle decomposition of $P$. Since $A_i^2=P$, we have $A_iP=PA_i$. Thus every $A_i$ commutes with $P$, and maps each cycle of $P$ only to another cycle of the same length. Hence the answer can be computed independently for each cycle length and multiplied.

Fix a length $d$, and let $m$ be the number of cycles of length $d$ in $P$. The action induced by the $K$ permutations $A_i$ on these cycles can be viewed as an action of $\mathbb F_2^K$. Therefore, each connected component contains exactly $2^r$ cycles for some $r$.

Let $c_{d,r}$ be the number of ways to build one component on a fixed set of $2^r$ cycles.

If $d$ is odd, then

$$c_{d,r} = {K \brack r}_2 (2^r-1)! d^{2^r-1}.$$

If $d$ is even, then $r=0$ is impossible, and for $r \ge 1$,

$$c_{d,r} = {K-1 \brack r-1}_2 2^{K-r} (2^r-1)! d^{2^r-1}.$$

Here

$${n \brack r}_2 = \prod_{i=0}^{r-1} \frac{2^{n-i}-1}{2^{r-i}-1}$$

is the Gaussian binomial coefficient over $\mathbb F_2$. Since $r \le \log_2 N$, it can be computed efficiently even for very large $K$.

Let $F[n]$ be the number of ways to process $n$ cycles of length $d$. If the component containing the first remaining cycle has size $2^r$, we choose the other $2^r-1$ cycles from the remaining $n-1$ cycles. Thus

$$F[0]=1,$$ $$F[n] = \sum_{2^r \le n} \binom{n-1}{2^r-1} c_{d,r} F[n-2^r].$$

Multiplying the values over all cycle lengths gives the answer. The total time complexity is $O(N\log N)$.