Because the axiom contains an additional condition: it’s “(K>0 and provability of ‘0=1’) implies 0=1”. Since you can’t prove K>0, you can’t apply Löb’s theorem.
Now, you could add an axiom “K=12” and get a new sound proof system which would prove PPT.2 consistent; or you could substitute 12 for every not-quoted occurrence of K in the axioms of PPT.2, and get a new sound proof system which would prove PPT.2 consistent; but that’s not a problem, because in both cases you get a new proof system, and there’s nothing unusual about being able to construct a new system in which the old system’s consistency can be shown.
Because the axiom contains an additional condition: it’s “(K>0 and provability of ‘0=1’) implies 0=1”. Since you can’t prove K>0, you can’t apply Löb’s theorem.
Now, you could add an axiom “K=12” and get a new sound proof system which would prove PPT.2 consistent; or you could substitute 12 for every not-quoted occurrence of K in the axioms of PPT.2, and get a new sound proof system which would prove PPT.2 consistent; but that’s not a problem, because in both cases you get a new proof system, and there’s nothing unusual about being able to construct a new system in which the old system’s consistency can be shown.