= Solution
In $A$, the tridentate ligand contains one anionic <phenoxide ligand>, a neutral secondary <amine> with its N–H bond intact, and a neutral <phosphine ligand>. The three <chloride ligands> contribute another three negative charges. Consequently the neutral complex has \b[Ti(IV), $d^0$, and 12 valence <electrons>]:
$$
\mathrm{OS(Ti)}=+4,\qquad \mathrm{VE}=0+2_{\mathrm O}+2_{\mathrm N}+2_{\mathrm P}+3(2)_{\mathrm{Cl}}=12.
$$
The N–H label is essential: counting this <nitrogen> as an anionic <amido ligand> would give the wrong <oxidation state>. A six-coordinate early-metal complex need not obey the <18-electron rule>.
In $B$, the substituted <cyclopentadienyl ligand> is $\eta^5$ and anionic, while the three-carbon bent ligand is an anionic $\eta^3$ <allyl ligand>, a four-electron donor in ionic counting. The pendant <pyridine> <nitrogen> is uncoordinated. Therefore \b[Mo(II), $d^4$, 18 <electrons>] follows from
$$
\mathrm{VE}=4_{d}+6_{\eta^5\mathrm{Cp}}+4_{\eta^3\mathrm{allyl}}+2(2)_{\mathrm{CO}}=18.
$$
The curved delocalization mark across the three-carbon ligand must not be read as a separate alkene or alkyne.
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