If A=0, B=1, and x=0 what are A' and B' in the logic circuit below?
A and B are D latches; on the clock edge, the value on the input is latched to the Q output.

A = 1 means that the Q output associated with the A latch is high (the complementary output is thus low).

A = 0 means that the Q output associated with the A latch is low (the complementary output is thus high).

A' is the value of A after the next clock edge.

This is a state machine.  A and B represent the current state.  The current state, along with the input x, determine the next state denoted by the primes on A and B.

So, for example, the input to the A latch is x so, if x = 0, on the next clock edge, 0 will be latched, i.e., the Q output of the A latch will be low; A' = 0.

Following the logic, the input to the B latch is:  \$(\bar x \cdot \bar A) + B \$

So, for x=0, A = 0, B = 0, the input to B is 1 and thus, after the next clock edge the Q output of B will be high; B' = 1. 

UPDATE:  a closer look shows that the schematic does not match the state transition table.  It's obvious that the input to the A latch is x and so, A' = x according to the schematic.  However, the 5th row in the state transition table has A' = 1 when x = 0.  That's not consistent.