Forget the words as whole units for a moment. This puzzle only cares about one thing: the vowels sitting inside each word, and a fixed number stamped on each vowel. Set the table first: $A=1$, $E=2$, $I=3$, $O=4$, $U=5$. Consonants are silent, they carry no weight at all, and the position of a vowel in the word does not matter either, only its identity and how many times it shows up.
Build a small table for the given pair before touching the answer choices:
The two scores tie at 10. That single fact is the whole rule of the game here, a word is "coded" into its partner when their vowel scores are equal. With the rule confirmed, apply it fresh to the second pair.
AUDIENCE breaks down into the vowels A, U, I, E, E:
\[ 1+5+3+2+2=13 \]So whatever option is correct, its vowels have to add up to exactly 13. Run each option through the same table:
Only one option survives the check, UNDERGONE, and it survives by hitting the target sum exactly, not approximately. Since the coding rule is an equality test and not a "closest match" test, there is no ambiguity once one option ties the score perfectly.
Let's summarize:
The word that AUDIENCE codes into is UNDERGONE.