The relation between solubility and K_(sp)Ksp is not a fixed one, but depends on the particular solid. Here's why
If the solid is of form AXAX, it will dissolve, producing one cation A^+A+ and one anion X^-X−.
AX(s) rightleftharpoons A^+ + X^-AX(s)⇌A++X−
and the K_(sp)Ksp expression is K_(sp)=[A^+][X^-]Ksp=[A+][X−]
If you let xx be the solubility, you can show that the ion concentrations will both be equal to xx, and the K_(sp)Ksp expression is given by x^2x2.
On the other hand, both A_2XA2X and AX_2AX2 dissolve to form three ions, and the K_(sp)Ksp expression is either
K_(sp)=[A^+]^2[X^-]Ksp=[A+]2[X−] or K_(sp)=[A^+][X^-]^2Ksp=[A+][X−]2
In either case, using the same assignment of xx equal to the solubility, you get to the same expression, the K_(sp)Ksp expression is equal to 4x^34x3.
Since all three solubilities are on the order of 10^(-4)10−4, the K_(sp)Ksp of AX will be on the order of 10^(-8)10−8, while the other two will be on the order of 10^(-12)10−12.
To break the tie, you look at the size of the numbers. AX_2AX2 at 1.20xx10^(-4)1.20×10−4 is smaller than 1.53xx10^(-4)1.53×10−4, and so, will generate the smaller K_(sp)Ksp value.