In the circuit shown below, the switch S is connected to position P...

When $S$ is connected to $P$,
In steady state,

$i = \dfrac{2 - 1}{2 + 1}$

$⇒ i = \dfrac{1}{3}A$

Now,

Charge on capacitor,

$q = CV$

$⇒ q = \left(1\mu F\right)\left(\dfrac{4}{3}V\right)$

$⇒ q = \dfrac{4}{3}\mu C$

$⇒ q = 1.33\mu C$