Number of permutation cycles in matrix transposition
Let's take a simple example:
$$ \begin{pmatrix} a &amp; b &amp; c \\ d &amp; e &amp; f \\ g &amp; h &amp; i \end{pmatrix}' = \begin{pmatrix} a &amp; d &amp; g \\ b &amp; e &amp; h \\ c &amp; f &amp; i \end{pmatrix}.$$
So we need to exchange $(b,d),(f,h),(c,g)$, i.e. the answer is 3. Now try to generalize this to $A \times B$ matrices. Obtain a formula, substitute $A = 2^a$ and $B = 2^b$, and find a way to efficiently compute the result modulo a small prime $p$.

When $A = 2^a$ and $B = 2^b$, things look nicer. The permutation taking a position in the original matrix to the corresponding position in the transposed matrix corresponds to left shift by an amount $a$. Suppose $g = \mathrm{gcd}(a,b)$ and $a' = a/g,b' = b/g$. We can now think of the operation as a left shift by the amount $a'$ of a string of length $a'+b'$ over an alphabet of size $G = 2^g$. The number of orbits is then the number of necklaces of length $a'+b'$, which has an explicit formula. (The shift amount doesn't matter since $a'$ is invertible modulo $a' + b'$ due to $(a',b')=1$.)