You can't really use pKa + pKb =14 for this problem. In reality, this equation is used to compared the conjugate acid to the conjugate base.
For example, assume you have RCOO-H <----> COO- + H+. I'm making up these numbers, but if the pKa of RCOO-H is 4 then the equation can be used to calculate the basicity strength of COO- (pKb=14-4=10). In other words, you cannot use the pKa of RCOO-H to calculate the pKb of RCOO-H, it could only be used for conjugates.
As far as this problem goes, I would suggest finding the dissociation constants.
-For A, the Ka would be around 1 x 10^-4 = [X-][H+]/[HX]
-Meanwhile D will have a Kb of around 1x10^-11 = [HX+][OH-]/[X]
The Ka tells you the strength of the acid, while the Kb tells you the strength of the base. The larger the Kb/Ka, the stronger the base/acid. In this case, D has a very, very small (10^-11) Kb so it's indicating that its a very weak base and most likely a stronger acid. Meanwhile, A has a Ka of 10^-4 and usually strong acids have a Ka > 1, so its most likely a weaker acid. So I'd most likely go with D.
In my opinion, I think this is a poor question, I don't think the test writers wrote it well since you cannot truly compare the acidity of molecules with different pKa/pKb, especially with monoprotic molecules.