Max Ionization Extent
For every reaction with an incomplete dissociation process in aqueous solution, you have to consider water's autoprotolysis.
\begin{aligned}\ce{
(1)&amp;&amp;HA + H2O &amp;&lt;=&gt; H3+O + A^{-}\\
(2)&amp;&amp;2H2O &amp;&lt;=&gt; H3+O + {}^{-}OH \\\hline
(3)&amp;&amp;HA + 3H2O &amp;&lt;=&gt; 2H3+O + A^{-} + {}^{-}OH
}\end{aligned}

The degree of dissociation is usually defined as
\begin{aligned}
&amp;&amp;\alpha &amp;= \frac{c(\ce{A^{-}})}{c_0}&amp;\text{with}&amp;&amp;c_0&amp;=c(\ce{A^{-}})+c(\ce{HA})\\
\therefore&amp;&amp;\alpha &amp;= \frac{c(\ce{A^{-}})}{c(\ce{A^{-}})+c(\ce{HA})}&amp;\text{with}&amp;&amp;
c(\ce{A^{-}})&amp;=K_a\frac{c(\ce{HA})}{c(\ce{H3+O})}\\
(4)\therefore&amp;&amp;\alpha &amp;= \frac{K_a}{K_a + c(\ce{H3+O})}
\end{aligned}

At infinite solution, the concentration of hydronium ions will be governed by the autoprotolysis $(2)$ and can therefore be rewritten as
\begin{aligned}
&amp;&amp;K_w&amp;=c(\ce{H3+O})c(\ce{{}^{-}OH})\\
(5)\therefore&amp;&amp;c(\ce{H3+O})&amp;=\sqrt{K_w}
\end{aligned}

At $25~^\circ{}C$ you will find your given equation:
\begin{aligned}
\alpha[\%] &amp;= \frac{K_a}{K_a + 1\cdot10^{-7}}\cdot100\%
\end{aligned}