= Solution
\b[The all-carbonyl <hydride>, HCo(CO)$_4$, is more acidic.] Loss of $\mathrm H^+$ gives $[\mathrm{Co(CO)_4}]^-$, whose negative charge is stabilized by <pi backbonding> into four <carbonyl ligands>. Replacing one CO by <triphenyl phosphite> reduces the net <electron>-withdrawing character of the ligand set relative to four carbonyls. The substituted <hydride> is more <electron> rich and its conjugate-base anion is less stabilized against <protonation>.
$$
\mathrm{HCo(CO)_3L}\rightleftharpoons\mathrm H^++[\mathrm{Co(CO)_3L}]^-,\qquad L=\mathrm{CO}\ \text{or}\ \mathrm{P(OPh)_3}.
$$
Both <conjugate bases> have a conventional 18-<electron> count, so the difference is not a missing-<electron> argument. It is the relative stabilization of acid and base by the <ligands in coordination chemistry>. This concerns heterolytic <deprotonation>, not the homolytic Co–H <bond-dissociation energy>.
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