How to calculate the pH of a buffer after adding HCl?
I'm not sure I follow your argumentation, but I guess you don't really need $K_\mathrm{b}$ at all since addition of strong acid ($\ce{HCl}$) influences the dissociation of the one buffer component - the weak acid ($\ce{NH4+}$):

$$\ce{NH4+ &lt;&lt;=&gt;[K_\mathrm{a}] NH3 + H+} \qquad K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H+}]}{[\ce{NH4+}]}$$

Henderson–Hasselbalch equation applied to this buffer system before the addition of acid allows to find initial $\mathrm{pH}$ (not required by the problem, I do this solely for demonstration):

\begin{align}
\mathrm{pH} &amp;= \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{NH3}]}{[\ce{NH4+}]}} \\
&amp;= -\log{(\pu{5.56e-10})} + \log{\frac{\pu{0.25 M}}{\pu{0.40 M}}} \\
&amp;= 9.05
\end{align}

Once the strong acid ($\ce{HCl}$, assuming complete dissociation) is added, the equilibrium shifts accordingly:

\begin{align}
\mathrm{pH_1} &amp;= \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{NH3}] - [\ce{HCl}]}{[\ce{NH4+}] + [\ce{HCl}]}} \\
&amp;= -\log{(\pu{5.56e-10})} + \log{\frac{\pu{0.25 M} - \pu{0.10 M}}{\pu{0.40 M} + \pu{0.10 M}}} \\
&amp;= 8.73
\end{align}

You would've needed $\mathrm{p}K_\mathrm{b}$ though when a strong base (e.g. $\ce{NaOH}$) were added.