Probably $U$ denotes union, and you mean $L = \\{ ba^nb^n \mid\ n\ge 0\\} \cup \\{ bba^nb^{2n} \mid\ n\ge 0\\}$.
Yes that can be done by a deterministic PDA. The first two letters of the string decide how to handle the remainder of the string. Pushing the $a$'s and popping the $b$'s in appropriate ratio is a standard task.