Initially $S_0$ is locked and it can be unlocked by either $P_1$ or $P_2.$ Say if $P_1$ proceeds first and then $P_2$ proceeds. In this case $S_0$ will be unlocked and process $P_0$ will print $’1’$ once. If process $P_0$ does wait(S0) before process $P_2$ does release $S_0$ it would have printed $1$ twice. So, correct answer here is at least $’1’$. Option D.