Magic square theory, in relation to any 3x3 array made up of nine consecutive numbers, states that the central cell will always be one third of the constant, so we can place the 9 with confidence.
Now, looking again at the original 1-to-9 basic square, we see that the eight perimeter numbers together sum to 40 (that is, eight times the central number).
Extrapolating this knowledge to our particular problem, it seems very probable that the eight cards around the perimeter will have to total to 8x9=72. For this to be so, we then find that precisely the set 5, 6, 7, 8, 9, 10, Jack, Queen, King will be needed. This leads swiftly to our answer, as shown here. You will notice that once again, the four 'odd' perimeter numbers must go in the middle of each row, as was the case with the 1-to-9 square.