Solution:

Let $D$ be the number of grid squares of which exactly one side belongs to $P$, let $E$ be the number of grid squares of which exactly two sides belong to $P$, which are moreover adjacent, and finally let $F$ be the number of grid squares of which exactly three sides belong to $P$. Since $m, n \geqslant 4$, among the $mn$ grid squares there is none whose entire boundary is contained in $P$, and consequently there are for the intersection of such a boundary with $P$ only the five possibilities addressed in the definitions of $B, C, D, E$ and $F$. We thus have
$$mn=B+C+D+E+F$$
Now we want to count in two ways the pairs $(Q, S)$ consisting of a grid square $Q$ and a side $S$ of $Q$ belonging to $P$. On the one hand, for the squares of type $B, C, D, E$ and $F$ there are respectively exactly $2, 0, 1, 2$ and $3$ possible choices for $S$, and hence this number amounts to $2B+D+2E+3F$. On the other hand, $P$ is composed of $(m-1)(n-1)$ unit segments, all of which can play the role of $S$, and each of these has two sides on each of which lies one grid square. Altogether we thus obtain
$$2(m-1)(n-1)=2B+D+2E+3F$$
At each of the $(m-1)(n-1)$ interior vertices of our rectangle, $P$ either passes straight through, and this happens exactly $A$ times, or $P$ turns. Whenever this happens, the two associated unit segments of $P$ enclose a grid square, and this must be either of type $E$ or of type $F$. Conversely, to each square of type $E$ belongs one, and to each square of type $F$ belong two, of these turning points of $P$. Altogether, therefore, we also have
$$(m-1)(n-1)=A+E+2F$$
From (2) $-$ (3) $+$ (1) we now obtain
$$m+n-1=A-B-C$$
and this in turn implies
$$A=B-C+m-n-1$$
With this the problem is solved.