I will assume that $A$, $B$ and $N$ are given, and that $d$ is unknown.
There is always a solution. Let $L$ be the perimeter of the ellipse. An obvious constraint is $N\,d. Take $d\in(0,L/N)$. As explained in Gerry Myerson's answer, pick a point $P_1$ on the ellipse, and then pick points $P_2,\dots,P_N$ such that $P_{i+1}$ is clockwise from $P_i$ and the euclidean distance between $P_i$ and $P_{i+1}$ is $d$. If $d$ is small, $P_N$ will be short of $P_1$, while if $d$ is large, it will "overpass" $P_1$. In any case, the position of $P_N$ is a continuous function of $d$. By continuity, there will be a value of $d$ such that $P_1=P_N$. It is also clear that this value is unique.
To find $P_N$ for a given $d$ you need to solve $N-1$ quadratic equations. To compute the value of $d$, you can use the bisection method.
Edit: TonyK's objections can be taken care of if $N=2\,M$ is even. Take $P_1=(A,0)$ and follow the procedure to find points $P_2,\dots,P_{M+1}$ in the upper semiellipse such that $P_{i+1}$ is clockwise of $P_i$ and at distance $d$, and $P_{M+1}=(-A,0)$. The the sought solution is $P_1,\dots,P_{M+1},\sigma(P_M),\dots,\sigma(P_2)$, where $\sigma(P)$ is the point symmetric of $P$ with respect to the axis $y=0$.
If $N=2\,M+1$ is odd, I believe that there is also a symmetric solution, but I have to think about it.