Phenolphthalein as a indicator for the titration of benzoic acid
You have to take two reactions into account.

First the protolysis of benzoic acid

$$\ce{HA + H2O &lt;=&gt; H3O+ + A-}$$
$$k_A = \frac{[\mathrm{H_3O^+}] [\mathrm{A^-}]}{[\mathrm{HA}]} \tag{1}$$

and second the (simplified) protolysis of phenolphthalein

$$\ce{HInd + H2O &lt;=&gt; H3O+ + Ind-}$$
$$k_{Ind} = \frac{[\mathrm{H_3O^+}] [\mathrm{Ind^-}]}{[\mathrm{HInd}]} \tag{2}$$

The color change is around the point where the concentrations $[\mathrm{HInd}]$ (colorless) and $[\mathrm{Ind^-}]$ (colored) are equal, i.e. $$k_{Ind} = [\mathrm{H_3O^+}] \tag{3}$$

If you substitute (3) into (1) and take the negative logarithm to base 10 you get

$$\mathrm{pK_A} - \mathrm{pK_{Ind}} = -\log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} = 4.4 - 9.4 = -5.2$$

what shows that the benzoic acid has been completely titrated at this point.