Over ZFA, the Parity Principle is strictly weaker. We'll show it follows from Multiple Choice, the assertion that for any family of nonempty sets $\mathcal{F},$ there is $g: \mathcal{F} \rightarrow [\bigcup \mathcal{F}]^{<\omega} \setminus \{\emptyset\}$ such that $g(S) \subseteq S$ for all $S \in \mathcal{F}.$ Multiple Choice holds in the second Fraenkel Model ($\mathcal{N}_2$ in *Consequences of the Axiom of Choice*), in which $\mathbf{C}_2$ fails badly (in particular, its atoms are a family of Russell socks).

Fix $X.$ Let $\sim$ be the equivalence relation on $\mathcal{P}(X)$ defined by $S \sim T$ if $S \triangle T \in [X]^{<\omega}.$ Let
$\mathcal{F}=\mathcal{P}(X) / \sim$ and let $g: \mathcal{F} \rightarrow [\bigcup \mathcal{F}]^{<\omega} \setminus \{\emptyset\}$ be such that $g(S) \subset S$ for all $S \in \mathcal{F}.$ Define a choice function
$h: \mathcal{F} \rightarrow \bigcup \mathcal{F}$ by $h(S)=\bigcup g(S).$ Then
$\mathcal{B}:=\{S \subseteq X: |S \triangle h([S]_{\sim} )| \in 2\mathbb{Z}\}$ is as desired.

Note that over ZF, Multiple Choice is equivalent to full Choice, so this approach doesn't immediately separate Parity Principle from $\mathbf{C}_2$ over ZF.