Simple way to extend AES to 256-bit block size
It turns out that this "mode" is distinguishable with two chosen messages; one in decrypt mode, and one in encrypt mode.

The first query is in decrypt mode, it is of the ciphertext $(C, 0)$ (where $C$ can be any nonzero value), and with an arbitrary IV.

Because of how decrypt works, $P_1$ of the resulting plaintext is $D_k(0) \oplus C$; this gives us the 128 bit value $Z = P_1 \oplus C$ with $E_k(Z) = 0$.

Our second query is in encrypt mode, it is of the plaintext $(Z, Z)$ with $IV=Z$.  If you go through how encryption works, we see that $C_0 = 0$, $C_1=0$, $C_2=0$, $C_3=0$, and so the resulting ciphertext is $(0,0)$.