(in progress...)
One thing should catch you eyes immediately. The first way creates $\ce{4 H+}$ for $\ce{1 [Cd(NH3)4]^2+}$, your way creates $\ce{1 H+}$ for $\ce{1 [Cd(NH3)4]^2+}$. And there can be formulated three other versions with $\ce{0 H+}$, $\ce{2 H+}$ and $\ce{3 H+}$ $\ce{H+}$ for $\ce{1 [Cd(NH3)4]^2+}$.
It is obvious maximally one of five versions can be true. And, in fact, not even one.
All five possible equilibrium constants $K_\text{eq} = K_\text{f} \cdot K_\text{a}^N$, $N=0..4$ for
$$\ce{Cd^2+(aq) + $(4-N)$ NH3(aq) + $N$ NH4+(aq) <=> [Cd(NH3)4]^2+(aq) + $N$ H+(aq)}$$
are formally correct and equilibrium concentrations will honor them in contexts of concentration = activity simplification.
BUT: The equilibrium equations do not say there is formed $\ce{$N$ H+}$ ions for each $\ce{[Cd(NH3)4]^2+}$ ion. It would have been true if concentration of $\ce{NH3}$ had been constant during the complex formation.
Let perform this formal substitution for easier expression manipulation:
$\ce{h = [\ce{H+}]}$, $\ce{n1 = [\ce{NH4+}]}$, $\ce{n2 = [\ce{NH3}]}$, $\ce{c1 = [\ce{Cd^2+}]}$, $\ce{c2 = [\ce{[Cd(NH3)4]^2+}]}$
and we can now formulate visually compact equations for equilibrii,
$$K_\text{a} = \frac{h \cdot n_2}{n_1}$$
$$K_\text{f} = \frac{c_2}{c_1 \cdot n_2^4}$$
for mass balances:
$$n_1 +n_2 + 4 c_2 = c_\text{N,tot}$$
$$c_1 + c_2 = c_\text{Cd,tot}$$
and for charge balance:
$$h + n_1 + 2 c_1 + 2 c_2 = c_\text{chrg}$$
$\ce{OH-}$ is neglected and furthermore due unknown temperature we do not know $K_\text{w}$.
$$K_\text{a} = \frac{h \cdot n_2}{n_1}$$
$$K_\text{f} = \frac{c_2}{c_1 \cdot n_2^4}$$
for mass balances:
$$n_1 +n_2 + 4 c_2 = c_\text{N,tot}$$
$$c_1 + c_2 = c_\text{Cd,tot}$$
and for charge balance:
$$h + n_1 + 2 c_1 + 2 c_2 = c_\text{chrg}$$