$A$ is finite, $B$ is NPC - When there's a polynomial reduction from $A$ to $B$?
I think you've confused the direction of the reductions. A reduction from $A$ to&nbsp;$B$ means that you could solve&nbsp;$A$ efficiently if you had an efficient way of solving&nbsp;$B$. It's certainly true that you could determine membership of a finite language efficiently if you could solve some NP-complete problem. After all, you can determine membership of a finite language efficiently already, without needing to solve, say, travelling salesman.

On the other hand, it would be surprising if there was a polynomial-time reduction from some NP-complete problem to the problem of determining membership of a finite language. That would be saying, "If I could determine membership of this finite language efficiently, I could solve every problem in NP efficiently."

In case you weren't confused about which way reductions go, here's a concrete reduction from the language $\mathcal{L} = \{\mathrm{elimination}, \mathrm{computer}, \mathrm{science}\}$ to Boolean satisfiability.  We will define a polynomial-time computable function $f$ from strings $\{a, \dots, z\}^*$ to Boolean formulae such that $f(w)$ is satisfiable if, and only if, $w\in \mathcal{L}$.

if w = "elimination" or w = "computer" or w = "science"
    return "X"
else
    return "X∧¬X"