Question about ε-moves at $\varepsilon NFA$ We learning know about ε-moves at $\varepsilon NFA$ and I'm little bit confuse...

If we have, for example, this  $\varepsilon NFA$:


and we have the word "$a$" - can we end the run at $s$? Or the running will be finish at $q$?
In other words what I'm tring to ask is this:
When we have an $\varepsilon$ - we first of all reading the letter from the input and then "jump"? or we can "jump" and then reading the input? (or we can do both).

I look at the definition of: $\hat{\delta}(q,a)$ (for a word) and $\delta(q,a)$ (for a letter), and I'm not sure if the answer to my question dependence at $\hat{\delta}(q,a)$ or $\delta(q,a)$, because:
$\hat{\delta}(q,a)=\{q,s\}$ and $\delta(q,a)=\{q\}$.

Thank you so much!
I hope I was clear (because I'm little bit confuse...)
In an $\varepsilon NFA$, when processing an $\varepsilon$ character, we can move on an $\varepsilon$-edge or we can stay on the state we are currently in. 
And for example since $a= \varepsilon a \varepsilon\varepsilon$, in your $NFA$ these two paths are both correct:

$$p \rightarrow^{\varepsilon} p \rightarrow^{a} q \rightarrow^{\varepsilon} s \rightarrow^{\varepsilon} s$$

and

$$p \rightarrow^{\varepsilon} p \rightarrow^{a} q \rightarrow^{\varepsilon} q \rightarrow^{\varepsilon} s$$

Therefore, with the character $a$ we have $\delta(p,a)=q$ and for the word $a$ we have $\hat{\delta}(p,a)=\{q,s\}.$