You're right, both ways work but I don't think that's what the problem is getting at. Or at least both ways work as long as the problem is really as narrow as having an event space with two binary, non-mutually-exclusive events (would be a stupid problem otherwise though, without more info).
I think the point of the problem is really just to think about the probability of being truthful when the event doesn't occur. That's the classic "bayes" logic example, 2 binary variables, 4 outcomes.
The four possibilities in this case are:
- $ T \wedge E $
- $ T \wedge \neg E $
- $ \neg T \wedge E $
- $ \neg T \wedge \neg E $
where "$\neg$" indicates that the event did not occur.
Regardless of which variable you condition on first, the probability of the witness being truthful is the same and you could express that truthfulness as:
$$p = P(E|T)P(T) + P(\neg E|\neg T)P(\neg T)$$
$$p = P(T|E)P(E) + P(\neg T|\neg E)P(\neg E).$$