If the sample's original pH is less than 8.3 (P alkalinity = 0) then most of the carbonic acid species will be bicarbonate ion and the pre-treatment alkalinity tells you how many moles of bicarbonate are present. For example, if you put 2 mmol NaHCO3 into distilled water there will be 2 mmol HCO3- ions, the pH will be 8.3, the P alkalinity will be 0 and the M alkalinity 2 mEq/L. If the bicarbonate are all replaced by (OH)- you will have 2 mmol/L OH- instead of 2 mmol/L bicarbonate and the pH would be 14 -(- log(2E-3)) = 11.3. The M alkalinity (the amount of acid required to neutralize all the (OH)-) is still 2 mmol/L but the P alkalinity, required to neutralize to pH 8.3 is
1000*(10^(11.3 - 14) - 10^(8.3 - 14)) = 1.993 (almost the same as the M alkalinity).
At pH 11 nearly all the carbonic acid species are carbonate but as all anions (bicarbonate and carbonate) have been replaced by (OH)- there is no bicarbonate left in the solution.
Now if you add 1 mmol NaCl and 1 mmol NaHCO3 to 1 L of water you should wind up with 1 mmol of (OH)- from replacement of Cl- ions and 1 mmol of (OH)- ions from replacement of HCO3- ions. You would, again, have a total of 2 mmol/L (OH-), a pH of 11.3 and P and M alkalinities that are almost equal and a chloride content of 0 or a very low value.
What was the original sample pH? How are you testing for chloride?