What is the final pH when 10 mL of 1M HCl is added to 1 L of 0.01 M of phosphate buffer at pH 7.2?

Phosphoric acid undergoes 3 ionizations with 3 distinct pKa values:

H3PO4 <--> H+ + H2PO4- pKa = 2.15

H2PO4- <--> H+ + HPO42- pKa = 7.20

HPO42- <--> H+ + PO43- pKa = 12.35

Because the pH of the buffer is equal to one of the pKa values, you know that in that buffer, [H2PO4- ] = [HPO42-] = 0.005 M.

Now, you are adding 0.010 L X 1.0 mol/L = 0.010 mol HCl

The first 0.005 mol of HCl will protonate the HPO42- in the solution and form a solution of just H2PO4-. The second 0.005 mol of HCl will protonate half of the H2PO4- forming 0.005 mol H3PO4 and leaving 0.005 mol of H2PO4-. Because you now have equal moles of H2PO4- and H3PO4, the solution is at the pKa of that first ionization. From the Henderson-Hasselbalch equation, the pH of the solution is:

pH = 2.15 + log (0.005 / 0.005) = 2.15