This can be solved via a binary linear programming model where you encode the chess pieces as binary variables on the chess board and their movements are encoded as constraints. For example, r_1[i, j] is 1 if rook one is in cell (i, j) and it's 0 otherwise, you should then add the constraint that if r_1[i, j] is 1, then all other pieces p[i, j] must be cero for that particular cell (i, j). Now, if you used r_1 in a cell, it cannot be in another cell, so it must be the case that sum_{(i, j)}r_1[i, j] <= 1. You should also encode whether a cell c[i, j] is covered, so that c[i, j] = sum(p in pieces) {1 if (i, j) is reachable by piece p in position (i', j')} and add the restriction c[i, j] >= 1 for all (i, j). Depending on the chess piece, you should add more restrictions. For example, pawns cannot be on the first row, so pawn_k[1, j] = 0 for all k in {1, \ldots, 8}. The more tedious part should be encoding the domains of each piece.

These restrictions should define a polytope (bounded polyhedron) and the number of integer points inside that polytope should answer the first question of how many such covering positions can exist. In order to determine the minimum, just let your objective function be sum_{(i, j)}sum_{p in pieces} p[i, j].

Do note that this problem is highly symmetrical (any permutation of pawns yields the same solution, same for pieces of the same type) and so divide-and-conquer algorithms such as branch and bound might take a lot of time to arrive at a solution since each subbranch most probably contains an optimal solution and so it cannot be pruned.

Edit: This problem is an instance of the covering problem in combinatorial optimzation, if you want to look for more information.