Correctness of Propagating Cipher Block Chaining mode of operation (PCBC)
How the PPCBC mode works
Propagating Cipher Block Chaining mode of operation (PCBC) works as with message indexes starts from 1;

For PCBC encryption we have;

$$C_i = E_K(P_i \oplus P_{i-1}\oplus C_{i-1})\text{ and } \color{red}{C_0 = P_0 \oplus IV}$$

For PCBC decryption ve have;

$$P_i = D_K(C_i) \oplus P_{i-1}\oplus C_{i-1}\text{ and } \color{red}{C_0 = P_0 \oplus IV}$$
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 $\color{red}{C_0 = P_0 \oplus IV}$