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For any line bundle $L$, such exact sequence splits. Indeed the map $L\rightarrow L$ induced by $p:L\oplus L\rightarrow L$ must be nonzero on one of the summands, say the first one; hence it is the multiplication by a nonzero scalar $\alpha $. Then the map $L\xrightarrow{\ (\alpha ^{-1},0)\ } L\oplus L$ is a section of $p$.