By <Tonelli theorem>,
$$
\chi(p)=\mathbb E_p|\mathcal C|
=\sum_{x\in\mathbb Z^d}\mathbb P_p(0\leftrightarrow x).
$$
If $p<p_c$, part (a) bounds the summand by $e^{-c\lVert x\rVert_\infty}$. There are only polynomially many vertices at each radius, so the series converges and $\chi(p)<\infty$.

Conversely, suppose $\chi(p)<\infty$. Then $p<1$. Choose $q>p$ so close to $p$ that
$$
2d\,\frac{q-p}{1-p}\,\chi(p)<1.
$$
Use the standard <sprinkling coupling for Bernoulli percolation>: first expose the $p$-open clusters, then independently open each remaining edge with probability $\alpha=(q-p)/(1-p)$. Explore the $q$-cluster of the origin cluster by following sprinkled edges. Each discovered $p$-cluster has at most $2d$ times its number of vertices as many incident edges, so the exploration is dominated by a <Galton-Watson process> of mean at most $2d\alpha\chi(p)<1$. This process dies out almost surely, and hence there is no infinite $q$-open cluster. Thus $q\leq p_c$, and $p<q$ implies $p<p_c$. Therefore
$$
\boxed{\chi(p)<\infty\iff p<p_c.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 214 / 2 / b / Solution

For $p<q$, first sample Bernoulli percolation at parameter $p$, then independently open each remaining closed coordinate with probability $(q-p)/(1-p)$. Every coordinate is then open with probability $q$, so the union has the parameter-$q$ law. This coupling separates existing $p$-clusters from the additional sprinkled connections between them.

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 Incoming links: Sprinkling coupling for Bernoulli percolation