A quick look at this problem reveals a few simple relations:
The pulley is experiencing forces both to the left and the right.
To the right it experiences tension from the string due to the block C pulling downwards due to gravity (I guess block C is suspended?) Let this tension be T. The value of tension can be obtained by $T = m_Cg$
To the left it experiences twice the tension from the string to the left due to the string dragging masses A and B along the table. Let this tension be $T'$. However the pulley experiences twice the tension as it is pulled by twice by the same string (eg: the string is looped around the pulley)
As the sum of forces on the pulley is
$F = T + 2T'$
...and since the pulley is massless $m = 0$.
Therefore, the sum of forces on the pulley $=0$ ($m = 0$ hence $F = ma = 0$).
...and from this we can say that $T = -2T'$ from our previous relation! (The negative sign simply implies that the tensions are acting in opposite directions)
Yes, the tension in the same string will be the same.
Now, since the only force acting on A is $T'$, acceleration of A can be calculated as:
$T' = m_Aa_A$
$a_A = T' / m_A$
Similarly with B:
$T' = m_Ba_B$
$a_B = T' / m_B$
As the acceleration of A and B vary due to their different masses, we can now say that they will move with different accelerations!