Proof of multiple equivalences.

The lack of symmetry in your proposal

• $P_1 \iff P_5$
• $P_2 \iff P_3$
• $P_1\land P_2\land P_3\land P_5 \iff P_4$

suggests that they are not jointly equivalent to the original

• $P_1 \implies P_2$
• $P_2 \implies P_3$
• $P_3 \implies P_4$
• $P_4\implies P_5$
• $P_5 \implies P_1.$

Rob has given a counter-interpretation; here's another one $$P_1 :=\; 7=7\\ P_5 :=\; 7=7\\ P_2 :=\; 8=9\\ P_3 :=\; 8=9\\ P_4 :=\; 8=9,$$ and yet another one $$P_2 :=\; 7=7\\ P_3 :=\; 7=7\\ P_1 :=\; 8=9\\ P_4 :=\; 8=9\\ P_5 :=\; 8=9.$$ The $5$ (non-equivalent) statements in each interpretation jointly satisfies the first (your proposal)—but not the second (the original)—set of sentences.

Here's an example set of sentences that is actually equivalent to the original set:

• $P_1\implies P_2\land P_3\land P_4\land P_5$
• $\lnot P_1\implies\lnot P_2\land \lnot P_3\land \lnot P_4\land \lnot P_5.$

(If, for example, $P_3$ is true, then by contraposition so is $P_1,$ and consequently so are the three remaining statements.)

Another that works is Hagen's third suggestion in their second comment under the OP.