Q:

How do I apply Thevenin theorem in this problem?

Can't find the answer to this question. Am I supposed to get the voltage in the 4 kilo ohm resistor? Really appreciate it if anyone can solve this one! Here's the circuit:

A:

There are plenty of ways to determine these values. I personally like superposition a lot because it is already part of the divide-and-conquer strategy promoted by the fast analytical techniques or FACTs. Here, superposition applies because we obviously deal with a linear circuit. We thus determine the voltage \$V_{th2}\$ across the 4-\$k\Omega\$ resistor when the 12-V source is turned off (a 0-V source is replaced by a short circuit) while the current source is alive. Then, we turn the current source off (a 0-A current source is an open circuit) and determine the voltage \$V_{th1}\$ when the 12-V is alive. At the end, we have \$V_{th}=V_{th1}+V_{th2}\$
We can now divide the circuit in sketches you can individually resolve:

In the upper sketch, we can use a Thévenin generator to determine the circulating current while the current source is off. Then, in the sketch below, shorting the voltage source simplifies the arrangement with two paralleled resistors. Use the current division property to obtain the voltage across \$R_3\$. If you enter your equations in a solver like Mathcad (see below) you find a Thévenin voltage of 6 V by summing both intermediate values:

To determine the Thévenin resistance, you can apply the FACTs because we have a linear circuit: turn all excitation sources off (12-V is shorted and the 2-mA is open circuited) then "look" at the resistance across the Thévenin terminals (across \$R_3\$). You see that this resistance is actually \$R_3\$ in parallel with some intermediate result the last sketch lets you find in a second, without an equation:
\$R_{int}=R_2||R_1+R_4\$ then bring \$R_3\$ back and you have \$R_{th}=R_3||(R_2||R_1+R_4)\$.
A quick SPICE simulation of the operating point confirms these results:

