Finding new pH after adding potassium hydroxide to citrate buffer
$\mathrm{p}K_\mathrm{a} = 3.06.$


Citric acid is a triprotic acid:

$$\ce{H3C6H5O7 &lt;=&gt;[K_1] H2C6H5O7- &lt;=&gt;[K_2] HC6H5O7^2- &lt;=&gt;[K_3] C6H5O7^3-}$$

The pKa given is for the second step. Citric acid is the fully protonated species, and citrate the fully deprotonated one. When you add the two, they neutralize to give the intermediate species.


  I'm trying to calculate it like initial $\mathrm{pH}$


The easiest way is to calculate the amount of protons available (3 x 10.26 mmol = 30.78 mmol) and the total amount of organic species (17.24 mmol). There would be no citrate (17.24 mmol of protons to protonate all with one proton, 13.54 mmol protons left), and no citric acid. 17.24 mmol - 13.54 mmol = 3.70 mmol would be monoprotonated and 13.54 mmol would be diprotonated. You can plug those numbers in (assuming that your volume is not crazy high or low) to get the initial pH.


  then if we add $\pu{4.72 mmol}$ $\ce{KOH},$ it's $1:3$ to citric acid.


No, because citrate is not a major species before or after adding the KOH. Instead, you would deprotonate 4.72 mmol of the diprotonated species to get 4.72 mmol more of the monoprotonated species.

$$\mathrm{pH_1} = 3.06 - \lg\frac{\pu{13.54 mmol} - \pu{4.72 mmol}}{\pu{3.70 mmol} + \pu{4.72 mmol}} = 3.04$$


  but this answer is incorrect. Where is the mistake? 


I think both answers are wrong, for the reasons given above.