Solution:

Let $ABCD$ have $n$ lattice points along the side $AB$. Then it has $kn$ lattice points along the side $AD$. Let $X$ be the first lattice point along $AB$ after leaving $A$. A shortest path from $X$ to $C$ must involve a total of $kn + n - 1$ moves between lattice points, $n - 1$ in the direction $AB$ and $kn$ in the direction $BC$. Hence the total number of such paths is
$$\frac{(kn + n - 1)!}{(kn)!\,(n - 1)!}$$
Similarly, the number of paths starting out along $AD$ is
$$\frac{(kn + n - 1)!}{(kn - 1)!\,n!}$$
Let $m = \frac{(kn + n - 1)!}{(kn - 1)!\,(n - 1)!}$. Then the number starting along $AB$ is $\frac{m}{kn}$ and the number starting along $AD$ is $\frac{m}{n}$, which is $k$ times larger, as required.