Probability of choosing link $L_1=p_1$
Probability for no bit error (for any single bit)$=(1-b_1)$
Similarly for link $L_2$
Probability of no bit error $=(1-b_2)$
Packet can go either through link $L_1$ or $L_2$ they are mutually exclusive events (means one event happens other won't be happening and so we can simply add their respective probabilities for the favorable case).
Probability packet will be received without any error = Probability of $L_1$ being chosen and no error in any of the $L$ bits + Probability of $L_2$ being chosen and no error in any of the $L$ bits
Hence, answer is option A .
Why option D is not correct choice here?
Option D here is giving the probability of a frame being arrived with at least one bit correctly - i.., all the bits are not errors.