Answer
Products: $NH_3$ and $H_2O$.
The right side will be favored.
Work Step by Step
$NH_4^+$ will donate one proton to $OH^-$
$NH_4^+(aq) + OH^-(aq) NH_3(aq) + H_2O(l)$
Since $NH_3$$(Kb = 1.8 \times 10 ^{-5})$ is stronger than $NH_4^+$$(Ka = 5.6 \times 10^{-10})$, the rigth side will be favored.