I claim $B$ wins. At any time, say the position on $B$'s turn is $(a,b)$ with $a\ge b$. We now have $3$ cases based on the value of $b\pmod{3}$: Case 1: $b\equiv 0 \pmod{3}$. Then, have $B$ eat pizzas until both have the same number. [You can show that it is not possible to have $a=b$ already in this case by induction.] Case 2: $b\equiv 1 \pmod{3}$. Then, have $B$ eat pizzas until $a=b+1$. If $a=b$, then have $B$ eat one from each pizzeria. [You can show that it is not possible for $a=b+1$ to already occur in this case by induction.] Case 3: $b\equiv 2 \pmod{3}$. Then, have $B$ eat pizzas until $a=b-1$. Since we have shown that it can never be the case that $a=b=0$ on $B$'s turn, then $B$ must eat the last pizza and thus win. $\blacksquare$ [\hide]