As depicted in the figure, two blocks of masses m and M are linked...

Let $a$ be the acceleration of the blocks and $\alpha$ be the angular acceleration of the pulley.

Since the string does not slip on the pulley, the relation between linear and angular acceleration is $a = \alpha r$.

Let $T_1$ be the tension in the horizontal segment of the string attached to mass $m$, and $T_2$ be the tension in the vertical segment attached to mass $M$.

Writing the equations of motion for each component:
For mass $M$ (moving downwards):
$Mg - T_2 = Ma$

For mass $m$ (moving horizontally on a smooth table):
$T_1 = ma$

For the pulley (rotating about its axis):
$(T_2 - T_1)r = I\alpha$

Substituting $\alpha = \dfrac{a}{r}$ into the torque equation:
$T_2 - T_1 = \dfrac{Ia}{r^2}$

Now, adding the equations for mass $M$, mass $m$, and the pulley:
$(Mg - T_2) + (T_2 - T_1) + T_1 = Ma + \dfrac{Ia}{r^2} + ma$

$Mg = a\left(M + m + \dfrac{I}{r^2}\right)$

Solving for $a$, we get:
$a = \dfrac{Mg}{M + m + \dfrac{I}{r^2}}$