Consider the arrangement displayed in the figure. The pulleys and...

Let $T$ be the tension in the string.

For the hanging mass $M$ moving downwards with acceleration $a$, the equation of motion is:
$Mg - T = Ma$

The movable pulley B is pulled towards the right by two segments of the string. The net force on the block of mass $2M$ is $2T$. Let its acceleration be $a_B$. The equation of motion is:
$2T = 2M a_B \Rightarrow T = M a_B$

Using the string constraint, if pulley B moves to the right by a distance $x$, both horizontal segments of the string shorten by $x$, releasing a total length of $2x$. Thus, mass $M$ moves downwards by $2x$.
Therefore, the acceleration of mass $M$ is twice that of mass $2M$:
$a = 2a_B \Rightarrow a_B = \dfrac{a}{2}$

Substituting $a_B$ into the tension equation:
$T = M \left(\dfrac{a}{2}\right) = \dfrac{Ma}{2}$

Substituting this value of $T$ into the first equation:
$Mg - \dfrac{Ma}{2} = Ma$

$Mg = Ma + \dfrac{Ma}{2}$

$Mg = \dfrac{3Ma}{2}$

$a = \dfrac{2g}{3}$