Concept:
A Finite Automaton accepts a string if, after reading the complete input string, it reaches a final (accepting) state.
In the given NFA:
• \(q_0\) is the start state.
• \(q_1\) is the accepting state.
• There is a self-loop on \(q_0\) labeled \(a,b\).
• There is a transition from \(q_0\) to \(q_1\) labeled \(a\).
The automaton can stay in \(q_0\) while reading any sequence of \(a\)'s and \(b\)'s and move to \(q_1\) only when it reads the final \(a\).
Step 1: Analyze the loop at \(q_0\).
The self-loop is labeled
\[
a,b
\]
which means the automaton may remain in \(q_0\) after reading either symbol.
Therefore, any prefix consisting of \(a\)'s and \(b\)'s can be processed while staying at \(q_0\).
Step 2: Analyze the transition to the final state.
There is a transition
\[
q_0 \xrightarrow{a} q_1.
\]
This means that the automaton can enter the accepting state only when an \(a\) is read.
Step 3: Determine which strings are accepted.
For a string to be accepted, the automaton must finish in the final state \(q_1\).
Since the transition into \(q_1\) is labeled \(a\), the last symbol of the accepted string must be
\[
a.
\]
Examples of accepted strings:
\[
a,\quad ba,\quad aba,\quad bbba,\quad aabaa.
\]
Step 4: Check the options.
The automaton does not require the substring \(ba\).
It does not require repeated \(ab\).
It simply requires that the last symbol be \(a\).
Hence the language is
\[
\boxed{\text{All strings ending with } a}
\]
Step 5: Write the final answer.
Therefore, the correct option is
\[
\boxed{\text{(C) All words that end with } a}
\]