= Solution
Write $u_k$ for the quantity denoted $\xi_k$ in the question, to distinguish it from the random-walk increments. Absorption at zero at time zero gives $u_0=1$, and starting at $N$ gives $u_N=0$. For $1\leq k\leq N-1$, condition on the first step and use independence of the remaining increments. That step contributes one factor of $s$, so the <discounted gambler's ruin hitting probability> satisfies
$$
\boxed{u_k=s(pu_{k+1}+qu_{k-1}),\qquad u_0=1,\quad u_N=0.}
$$
The absorption time is finite almost surely: in each block of $N$ steps there is a probability bounded below by a positive constant of reaching a boundary by successive steps in one direction.
Trying $u_k=r^k$ gives $psr^2-r+qs=0$, with roots
$$
r_\pm=\frac{1\pm\sqrt{1-4pqs^2}}{2ps}.
$$
They are positive and distinct because $4pq\leq1$ and $0<s<1$. Hence the general solution is $Ar_+^k+Br_-^k$. Enforcing both boundary values gives
$$
\boxed{\xi_k=u_k=\frac{r_+^Nr_-^k-r_-^Nr_+^k}{r_+^N-r_-^N},\qquad0\leq k\leq N.}
$$
This includes the unbiased walk without a special limiting case, since the discount keeps the roots distinct. For uniqueness, a difference of two solutions has zero boundary values and its largest absolute value $M$ satisfies $M\leq sM$ if attained in the interior. Since $s<1$, $M=0$.
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