The relation between solubility and K_(sp) is not a fixed one, but depends on the particular solid. Here's why
If the solid is of form AX, it will dissolve, producing one cation A^+ and one anion X^-.
AX(s) rightleftharpoons A^+ + X^-
and the K_(sp) expression is K_(sp)=[A^+][X^-]
If you let x be the solubility, you can show that the ion concentrations will both be equal to x, and the K_(sp) expression is given by x^2.
On the other hand, both A_2X and AX_2 dissolve to form three ions, and the K_(sp) expression is either
K_(sp)=[A^+]^2[X^-] or K_(sp)=[A^+][X^-]^2
In either case, using the same assignment of x equal to the solubility, you get to the same expression, the K_(sp) expression is equal to 4x^3.
Since all three solubilities are on the order of 10^(-4), the K_(sp) of AX will be on the order of 10^(-8), while the other two will be on the order of 10^(-12).
To break the tie, you look at the size of the numbers. AX_2 at 1.20xx10^(-4) is smaller than 1.53xx10^(-4), and so, will generate the smaller K_(sp) value.
