Let T be an event whose occurrence is to be investigated and the probability with which a witness(W) tells the truth is $p$. Let E be the event of witness testifying that the event T has occurred, then how can we express $p$ in term of T and E?
In my view expressing $p$ as $Pr[T|E]$ should be the correct way of expressing this probability but I cannot find any fault if we express $p$ as $Pr[E|T]$. Some insight into it would be useful.
On the one hand, your question may be interpreted as a question related to the language. What does the following sentence mean?
$$\color{red}{\text{The witness tells the truth.}}$$
Note that we examine the witness' truth-telling potential only in case when the event $T$ does happen. (This is immediately questionable.) However, let's accept the OP's approach first.
In this case there are two interpretations (as hinted by the OP)
1.
Assuming that $T$ has happened, the witness tells that $T$ has happened (This is $E$.). Now, $$p_1=P(E\mid T).$$
2.
Assuming that the witness tells that $T$ has happened ($E$ occurs), $T$ did happen. Now,
$$p_2=P(T\mid E).$$
There is a more exact (statistical) approach. Assume that we can repeat the experiment (independently and many times) with the witness and the event. So, we have the following table.
$$ \begin{matrix} \text{No.}& \text{the event has happened}& \text{the witness says that the event has happened}\\ 1&\text{yes}&\text{yes}\\ 2&\text{no}&\text{yes}\\ 3&\text{yes}&\text{yes}\\ 4&\text{no}&\text{yes}\\ 5&\text{yes}&\text{no}\\ 6&\text{yes}&\text{yes}\\ \vdots\\ 100&\text{yes}&\text{no}\\ 101&\text{no}&\text{yes}\\ 102&\text{yes}&\text{yes}\\ 103&\text{no}&\text{yes}\\ 104&\text{no}&\text{no}\\ \vdots \end{matrix} $$
ad 1.
We can approximate $P(E\mid T)$ the following way. We count the rows in the table where $T=\text{yes}$. Let's denote this number by $N_T$. Considering the visible rows $$N_T=6.$$
Then, among these rows count those when the witness tells yes, the truth. Let this number be denoted by $N_{T,E}$. Considering the visible rows
$$N_{T,E}=4.$$
The approximation of the probability that the witness tells the truth is
$$p_1=P(E\mid T)\approx\frac46=\frac23 .$$
ad 2.
We can approximate $p_2=P(T\mid E)$ the following way. We count the rows in the table where $E=\text{yes}$. Let's denote this number by $N_E$. Considering the visible rows $$N_E=8.$$
Among these rows, now, let's count those where $T=\text{yes}$. Notice that this number is $N_{T,E}=4$ again. Our approximation is
$$p_2=P(T\mid E)\approx\frac12.$$
The two interpretations seem to be contradicting.
What might be the cause behind this contradiction? Remember we've been considering the witness telling the truth only if the event did occur and the witness told so. This is not a good definition of telling the truth.
So let's use the following definition $$\color{green}{\text{The witness tells the truth}}$$
if he says "yes" if $T$ did occur, and he says "no" if $T$ did not occur. Using our notation the probability of telling the truth in the latter sense
$$\color{green}{p=P(E\cap T \cup E^c \cap T^c)=P(E\cap T)+P(E^c \cap T^c)}.$$
Considering the visible part of our table we have $N=11$ cases. The number of cases when $T$ and $E$ takes place together, $N_{T,E}=4$. The number of cases when two no's appear together is $1$. So,
$$p=\frac4{11}+\frac1{11}=\frac5{11}.$$
Our contradiction might have been caused by the two possibilities of writing $P(E\cap T)$ two different ways:
$$P(E\cap T)=\color{blue}{P(T\mid E)P(E)}=\color{purple}{P(E\mid T)P(T)}.$$
Even if $P(T\mid E)\not =P(E\mid T)$
$$P(T\mid E)P(E)=P(E\mid T)P(T).$$
Based on the visible part of the table $P(E)\approx \frac8{11}$ and $P(T)\approx \frac6{11}.$
So, $$P(T\mid E)P(E)\approx\frac12\cdot \frac8{11}=\frac4{11}$$ and $$P(E\mid T)P(T)\approx \frac23 \cdot \frac 6{11}=\frac4{11}.$$
Conclusion
Neither $P(E\mid T)$ nor $P(T \mid E)$ are suitable "translations" of the sentence $$\color{green}{\text{The witness tells the truth}}$$
because both formulas are influenced by the probabilities of the events in the condition. Furthermore, the influence of these (a priory) probabilities is not symmetric.
The correct "translation" is the symmetric formula
$$\color{green}{p=P(E\cap T)+P(E^c \cap T^c)}.$$