I was looking at the Wikipedia article about Block cipher mode of operation and I was wondering how can you from the encryption $C_i = E_K(P_i \oplus P_{i-1}\oplus C_{i-1})$ get the decryption $P_i = D_K(C_i) \oplus P_{i-1}\oplus C_{i-1}$ ?
Is there even a way to get decryption from encryption (and vice versa) or not?
How the PPCBC mode works
Propagating Cipher Block Chaining mode of operation (PCBC) works as with message indexes starts from 1;
$$C_i = E_K(P_i \oplus P_{i-1}\oplus C_{i-1})\text{ and } \color{red}{C_0 = P_0 \oplus IV}$$
$$P_i = D_K(C_i) \oplus P_{i-1}\oplus C_{i-1}\text{ and } \color{red}{C_0 = P_0 \oplus IV}$$
given $C_1$ to decrypt $P_1$
$$P_1 = D_K(C_1) \oplus P_{0}\oplus C_{0} = P_1$$
given $C_2$ to decrypt $P_2$
$$P_2 = D_K(C_2) \oplus P_{1} \oplus C_{1}$$ Now we can solce since we know, $P_1$ and we have all $C_i$s.
and so on...
The red part was missing in your question.
The Correctness of the PCBC mode
An encryption scheme must satisfy the correctness requirement; for every key $k$ output by the key generation algorithm and for every message $m \in \mathcal{M}$ ($\mathcal{M}$ is the message space), the following must hold;
$$ D_k(E_k(m)) = m$$
To see that PCBC has the correctness; take
$$C_i = E_K(P_i \oplus P_{i-1}\oplus C_{i-1})$$ take decryption on the both sides
$$D_K(C_i)= \color{red}{D_K}(\color{red}{E_K}(P_i \oplus P_{i-1}\oplus C_{i-1}))$$ cancel Enc with Dec.
$$D_K(C_i)= P_i \oplus P_{i-1}\oplus C_{i-1}$$
Therefore:
$$P_i = D_K(C_i) \oplus P_{i-1}\oplus C_{i-1}$$
Since we know, all $C_{i}$s, and previously decrypted $P_{i-1}$.
If it is the first case $(i=1)$, then we already know $\color{red}{C_0 = P_0 \oplus IV}$
