Cipher Block Chaining
The problem suggests that the encryption does not use padding, and $C_0=\mathit{IV}$, $C_0'=\mathit{IV}'$. Then
$$E_k(C_0\oplus P_1) = C_1$$
$$E_k(C_1\oplus P_2) = C_2$$
$$E_k(C_0'\oplus P_1') = C_1'$$
You also have
$$C_1'=C_2$$
That is enough to find $P_1'$.