Acceptance by final state and by empty stack are equivalent for PDAs

For final-state to empty-stack, adjoin a fresh bottom marker $B$ below the original stack and a fresh drain state $d$. Whenever the simulated machine reaches an accepting state after consuming all input, the new machine takes an $\epsilon$-move to $d$ and in state $d$ repeatedly pops the entire remaining stack, including $B$. By [L1] and [L2], the new machine accepts exactly the same words, now by empty stack.

For empty-stack to final-state, adjoin a fresh start state $s$, a fresh bottom marker $B$, and a fresh accepting state $f$. The new machine first pushes the old initial symbol above $B$, then simulates $M$, and the only way to enter $f$ is by an $\epsilon$-move that pops $B$ after the simulated stack has become empty and the input has been fully consumed. By [L1] and [L2], this converts empty-stack acceptance into equivalent final-state acceptance.

The two constructions from steps 1.1 and 1.2 prove both directions of the equivalence.