Does this block cipher mode allow for decryption? Is there's a way for someone (with the key) to decrypt a message encrypted with the cipher mode shown?



$$ P_0 = IV $$
$$ C_i = P_{i-1} \oplus E_K(P_i) \oplus P_i $$
Ok, I will assume that:


The block you have labeled as "Encryption" is some sort of block cipher
You run this mode to transform the IV and the plaintext into a ciphertext
You are asking whether someone with the IV, the ciphertext and the block cipher key can recover the original plaintext.


Well, the quick answer is that they can't.  This is true in two senses:


It is difficult to find a message that would encrypt to the given ciphertext.  To recover (say) the first plaintext block, they would need to find a $P_1$ that solves the equation:

$C_1 \oplus IV = E_K( P_1 ) \oplus P_1$

However, for conventional block ciphers, there's no easy way to solve for such a $P_1$.  Now, if the receiver knew the last plaintext block, he could decrypt all the rest.  However, that isn't a standard assumption.
There might be multiple plaintext messages that encrypt into the same value.  This is because, while $E_K( P_i )$ is a bijection (this is a standard assumption of block ciphers), there's no reason to suppose that $E_K( P_i ) \oplus P_i $ is (and, in fact, with high probability, it isn't).


To give a simple example of this last point, why don't see assume that we have a 1 byte block cipher with:

$E_k( 0x00 ) = 0x71$

$E_k( 0x17 ) = 0x66$

Then, if $IV = 0x19$, then:

The encryption of the single byte message $0x00$ will be the single byte $0x19 \oplus E_k( 0x00 ) \oplus 0x00 = 0x19 \oplus 0x71 \oplus 0x00 = 0x68$

The encryption of the single byte message $0x17$ will be the single byte $0x19 \oplus E_k( 0x17 ) \oplus 0x17 = 0x19 \oplus 0x66 \oplus 0x17 = 0x68$

Hence, in this case, if we get an encrypted message $0x68$, we have no idea if the original plaintext was $0x00$ or $0x17$