Video solution:

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Written solution:

Firstly, note that $$\text{FLYFLY} = 1001(\text{FLY})$$ and, similarly, $$\text{BUGBUG}=1001(\text{BUG})$$ so the equation can be simplified to $8 \cdot \text{FLY} = \text{BUG}.$

For $\text{BUG}$ to remain three digits, $F$ must be $1.$ Moreover, $L$ must also be less than $3$ to avoid carrying over $2$ to the hundreds digit and making the product $4$ digits. Since we need $\text{FLY}$ to be the greatest number, $L$ must be $2.$

To identify the possible values for $Y,$ we note that so far we have $8(12)=96$, so we must avoid carrying $4$ to the tens digit to keep the resulting product three digits. Hence, $Y\le5$. We can try $4$ and verify that the resulting product has unique digits that haven't been used yet: $8(124) = 992$, which does not have unique digits. Trying $3,$ we get $8(123) = 984$, which satisfies our criteria.

Hence, $$\text{FLY} + \text{BUG} = 123 + 984 = 1107.$$

Thus, C is the correct answer.