Use a bounded Bellman-Ford style dynamic programming approach.

With at most k stops, the path may use at most k + 1 flights. Repeatedly relax all flights, but each round must use a copy of the previous distances so that one round adds only one more flight.

After k + 1 rounds, the destination distance is the cheapest valid price.

Pseudocode:

function cheapestFlightWithinKStops(nodes, flights, size, source, destination, stops):
    distance = array filled with infinity
    distance[source] = 0
    repeat stops + 1 times:
        nextDistance = copy of distance
        for each flight [from, to, price]:
            if distance[from] is not infinity:
                nextDistance[to] = min(nextDistance[to], distance[from] + price)
        distance = nextDistance
    if distance[destination] is infinity:
        return -1
    return distance[destination]