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.
