Equilibrium, Kp constant, Partial Pressures A system in equilibrium contains 
$$I_{2_(g)}\ \text{and} \ I_{(g)}$$
under pressures $$P_1=0,21 \,\text{atm. and}\  P_2=0,23 \,\text{atm.}$$
While its temperature remains constant, we reduce the system's volume to half its original value.

I have to determine $P_1'$ when the system's equilibrium is reestablished.

I know I can calculate $K_p$ and I know I somehow have to use $PV =NRT$, but from there on I am lost, I cannot figure out how to come up with enough equations to find all of my unknowns. Can anyone show me an explicit solution to the problem?

What I tried, but got really nowhere:

$$2I_{(g)} \leftrightarrow I_{2_{(g)}}, K_p=\frac{P_1}{P_2^2}=3,97.$$
$$P_t=P_1 + P_2 = 0,44 \,\text{atm.}$$

When the new equilibrium is established :

$$Kp=\frac{P_1'}{P_2'^2}=3,97$$

$$\frac{2P_1 }{P_1'}=\frac{n_1}{n_1'}$$
and same with $P_2, P_2', n_2, n_2'$, but I don't know where to go from there.
The method referred to in the comment is not correct and is a common mistake. One may think,

$$p_1'+p_2'=p_t\frac{V_t}{V_t'}=2p_t=0.88 \rm atm,$$

but this is invalid since $n_1+n_2\neq n_1'+n_2'$, thus $p_tV_t\neq p_t'V_t'$

The correct way is to use the fact that the number of iodine atoms is constant, from which we get,
$$2n_1+n_2=2n_1'+n_2',$$
$$(2p_1+p_2)V_t=(2p_1'+p_2')V_t',$$
$$1.3 {\rm atm}=4p_1+2p_2=2p_1'+p_2',$$

Then you can go on by combining this with the equation you have already obtained from $K_p$ (listed in your question) and solve for $p_1'$ and $p_2'$.