What is certain is that KOH + pAPH = KpAP + HOH which is a neutralization.
But, KpAP + HOH is an equillibrium reforming KOH + pAPH which is present as K + OH. Also, if you look at the molar equivalents used above, 100 g of NaOH in this case is over 2 moles of hydroxide and less than 1 mole of pAP. Even if it were the sulfate the pAP would be overwhelmed by the caustic. The Metabisulfite contributes little to this reaction in terms of OH, merely introducing the SO3= ion.
As a result, regardless of what is said, if the pH is above 7, the solution is alkaline and contains OH ions. This is fundamental chemistry. If it were acidic, then there would be H+ ions. At 7.0, there would be equal numbers of H+ and OH-. Thats fundamentally just water. That is the only place on the pH scale where there is no excess of acid or base, and that is called neutrality.
The caustic may even be equimolar to pAP or its salt, but the equillibrium causes an excess of OH- to form in the solution and that is a chemical fact. That is why the pH is so high, due to an excess of OH-. Dr Andresen may have been in error or trying to mislead people, IDK, but his statement is certainly misleading in the face of the formulas and pH values I have seen.