= Solution
Differentiate the normalized average, including the <derivative> of its denominator. The <enthalpy fluctuation formula> follows from
$$
\partial_\beta\langle H\rangle_P
=-\langle H^2\rangle_P+\langle H\rangle_P^2
=-\operatorname{Var}(H).
$$
Since $d\beta/dT=-1/(k_BT^2)$,
$$
\boxed{C_P=\frac{\operatorname{Var}(H)}{k_BT^2}}.
$$
Thus a larger equilibrium spread of <enthalpy> corresponds to a larger response to heating at fixed <pressure>. The <variance> is nonnegative, so this finite-system equilibrium description has $C_P\geq0$. The relation connects a response coefficient to fluctuations of the appropriate exchanged quantity; at fixed <volume> the corresponding quantity is <internal energy>.
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