How do the pKa and pH indicate to the proper environment of protonation?
The sentence immediately before the bolded one indicates that the $\mathrm pK_\mathrm a$ value of the ligand bound to the enzyme was determined to be $9.1$. It also indicates that the $\mathrm pK_\mathrm a$ value o the free ligand is 2.5 logarithmic units less, i.e. $\mathrm pK_\mathrm a(\ce{Lig}) \approx 6.6$.

The human cell is well-buffered to ensure that the pH stays within a very narrow range. There are enough buffering compounds around, that we never need to use any Henderson-Hasselbalch equation or other difficult calculation methods to determine the protonation state of a given compound. We can simply examine its $\mathrm pK_\mathrm a$ value; if that is lower than the cell’s pH value, the compound is deprotonated. In all other cases, it is not.

The $\mathrm pK_\mathrm a$ value of the free ligand indicates it should be deprotonated in the cell. However, upon binding to the protein with the $\mathrm pK_\mathrm a$ value rising by $2.5$  logarithmic units, the ligand suddenly becomes protonated. This is what the bolded sentence implies.