As I understand the problem it is identical to the Monty Hall problem if you map survival-prize and doors-prisoners. The opposite of survival is death and the opposite of prize is nothing.

So when the guard says B is sure to die he is actually opening door number 2 to show no prize is there.

Before this event the odds of survival for each prisoner were 1/3, 1/3, 1/3.

The odds that the guard would choose prisoner B as the non-surviving one (dead that is) in case prisoner A/B/C is the surviving one are: 1/2, 0, 1.

That is if A was pardoned then the guard can chose either of B or C so the odds of pointing out B are 1/2. The odds of pointing out B as dead when he is pardoned are 0 and the odds of pointing out B as dead when C is pardoned are 1 (because he cannot indicate A as dead because the guard is not allowed to tell A about his own fate). Hence the final odds for each prisoner must add up to 1 so through proportionality they are 1/3, 0, 2/3.

So prisoner A didn't find out anything new about his chance of being the pardoned one.