 Solution

First give $w$ the vote of every voter whose allowed set contains $w$. Let their number be $K$. This loses no possible strict victory: starting with any winning assignment, changing such a ballot to $w$ increases $w$'s score and weakly decreases an opponent's score. Thus a winning assignment exists exactly when one exists with these $K$ ballots fixed.

Every remaining voter must now be assigned to an allowed opponent, and each opponent may receive at most $K-1$ votes. If any allowed set is empty, there is no valid election assignment. With at least two candidates, $K=0$ immediately makes a unique victory impossible. With only the candidate $w$, every nonempty ballot set forces that candidate, so the answer is yes under the question's strict-comparison definition.

For the remaining case $K\ge 1$, make a flow network with a vertex for each remaining voter and each opponent. Add source-to-voter arcs of flow network edge capacity one, voter-to-opponent arcs of flow network edge capacity one for the allowed choices, and opponent-to-sink arcs of flow network edge capacity $K-1$. Compute an integral maximum flow. ThenA flow saturating all voter arcs assigns each remaining voter to exactly one allowed opponent, by integrality and flow conservation. The sink-arc capacity constraints give every opponent strictly fewer than $K$ votes. Conversely, any winning assignment with the fixed $K$ votes defines just such a saturating feasible flow. The flow network has at most $n+m+2$ vertices and $O(n+m+{\sum }_i|C_i|)$ arcs, hence the decision takes polynomial time.