Dissociation constant for Indicators
First of all, the indicator does not change colour instantaneously at $\mathrm{pH = pK_a}$. The indicator is a different colour than it's conjugate base, and exists in an equilibrium with it in solution. For the colour change to be noticed, the concentration of the base has to be at least 10 times more than that of the conjugate acid, or vice versa. This is why there is a range of colour change(look at these pH charts), with the upper and lower bounds having a $\mathrm{pH}$ difference of $2$. As an example, here's a simplified reaction for Phenolphthalein ($\ce{Ph}$ is the phenolphthalein anion)
$$\ce{\underset{\text{colourless}}{H2Ph} &lt;=&gt; 2H+ + \underset{\text{violet}}{Ph^2-}}$$
If you take a simpler, weak monobasic acid indicator $\ce{HIn &lt;=&gt; H+ + In-}$, the $\mathrm{K_a}$ is
$$\mathrm{K_a} = \frac{\ce{[H+][In^-]}}{\ce{[HIn]}}$$
The relative concentration of the indicator and it's conjugate base is only determined by the concentration of the $\ce{H+}$ ion, as $\mathrm{K_a}$ is a constant.
Answering your second query, the author takes the condition of equilibrium, simply because the system is in equilibrium, that is $\mathrm{Q_a = K_a}$. It is assumed that the reaction is instantaneous or that titration is performed slowly enough so that the system is in equilibrium at every single instant.