Aspartic acid conjugate base acidity
A simple way to compare the $\mathrm pK_\mathrm a$ of $\ce{-NH3+}$ and $\ce{-COOH}$ in such a scenario would be to assume the situation when the amino acid becomes a zwitter-ion.
A zwitter-ion is a molecule that has both positive and negative charges on it. In an amino acid, the $\mathrm {pH}$ at which this is seen is known as the isoelectronic point.



Let's consider this molecule to have reached the isoelectronic point. We know that the value would be less than $7$ since there are two acidic groups (meaning that this is an acidic amino acid). When we reach that point, we see that the both the carboxyl groups are de-protonated and the amine group still exists as $\ce{-NH3+}$.
This means that the $\ce{-NH3+}$ group is less acidic than the the acid groups. Therefore, Hd is is less acidic than Ha or Hb.
As you further increase the $\mathrm{pH}$, we see that at a point, the $\ce{-NH3+}$ group becomes $\ce{-NH2}$, before Hc is removed. This means that Hd is more acidic than Hc.
Now, we need to compare the two acid groups. As you may have noticed, The $\ce{-NH3+}$ group is closer to Hb than Ha, since this group is a highly electron withdrawing group [-I effect], the negative charge formed by the de-protonation of Hb would be more effectively withdrawn making it more stable than the conjugate base formed by the de-protonation of Ha.
Therefore the order of stability would be:

Hb &gt; Ha &gt; Hd &gt; Hc