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 $\ce{3 H+}$ 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. Before introduction of the cadmium salt, there is established an equlibriumthe equilibrium ammonia concentration and thisdue ammonium hydrolysis. This concentration shifts when the cadmium complex is formed.
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 $K_\text{w}$.
Substituting $c_2 = c_\text{Cd,tot} - c_1$:
$$K_\text{a} = \frac{h \cdot n_2}{n_1}$$ $$K_\text{f} = \frac{c_\text{Cd,tot} - c_1}{c_1 \cdot n_2^4}$$ $$n_1 +n_2 + 4 (c_\text{Cd,tot} - c_1) = c_\text{N,tot}$$ $$h + n_1 + 2 c_1 + 2 (c_\text{Cd,tot} - c_1) = c_\text{chrg}$$
Substituting $n_1 = c_\text{chrg} - 2c_1 - 2(c_\text{Cd,tot} - c_1) - h$:
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{chrg} - 2 c_1 - 2 (c_\text{Cd,tot} - c_1) - h}$$ $$K_\text{f} = \frac{c_\text{Cd,tot} - c_1}{c_1 \cdot n_2^4}$$ $$(c_\text{chrg} - 2c_1 - 2(c_\text{Cd,tot} - c_1) - h) +n_2 + 4 (c_\text{Cd,tot} - c_1) = c_\text{N,tot}$$
Simplifying:
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{chrg} - 2 c_\text{Cd,tot}- h}$$ $$K_\text{f} = \frac{c_\text{Cd,tot} - c_1}{c_1 \cdot n_2^4}$$ $$c_\text{chrg} + 2c_\text{Cd,tot} - h +n_2 - 4c_1 = c_\text{N,tot}$$
Substituting $c_1 = \frac 14(c_\text{chrg} + 2c_\text{Cd,tot} - h +n_2 - c_\text{N,tot})$
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{chrg} - 2 c_\text{Cd,tot}- h}$$ $$K_\text{f} = \frac{c_\text{Cd,tot} - (\frac 14(c_\text{chrg} + 2c_\text{Cd,tot} - h +n_2 - c_\text{N,tot}))}{(\frac 14(c_\text{chrg} + 2c_\text{Cd,tot} - h +n_2 - c_\text{N,tot})) \cdot n_2^4}$$
Simplifying:
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{chrg} - 2 c_\text{Cd,tot}- h}$$ $$K_\text{f} = \frac{2c_\text{Cd,tot} - (c_\text{chrg} - h +n_2 - c_\text{N,tot})} {(c_\text{chrg} + 2c_\text{Cd,tot} - h +n_2 - c_\text{N,tot}) \cdot n_2^4}$$
Simplifying: As $c_\text{chrg} = 2c_\text{Cd,tot} +c_\text{N,tot}$:
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{N,tot}- h}$$ $$K_\text{f} = \frac{ h - n_2 } {(c_\text{chrg} + 2c_\text{Cd,tot} - h +n_2 - c_\text{N,tot}) \cdot n_2^4}$$
Approximating: $c_\text{N,tot} \gg h$ and $c_\text{chrg} + 2c_\text{Cd,tot} - c_\text{N,tot} \gg |- h + n_2| $
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{N,tot}}$$ $$K_\text{f} = \frac{ h - n_2} {(c_\text{chrg} + 2c_\text{Cd,tot} - c_\text{N,tot}) \cdot n_2^4}$$
Substituting: $n_2 = \frac{K_\text{a} \cdot c_\text{N,tot} }{ h}$
$$K_\text{f} = \frac{ h - (\frac{K_\text{a} \cdot c_\text{N,tot} }{ h})} {(c_\text{chrg} + 2c_\text{Cd,tot} - c_\text{N,tot}) \cdot {(\frac{K_\text{a} \cdot c_\text{N,tot} }{ h})}^4}$$
$$K_\text{f} = h^3\frac{ h^2 - ({K_\text{a} \cdot c_\text{N,tot} })} {(c_\text{chrg} + 2c_\text{Cd,tot} - c_\text{N,tot}) \cdot {({K_\text{a} \cdot c_\text{N,tot} })}^4}$$
The Excel based iteration $h=\mathrm{f}(h)$ provided
$$h=[\ce{H+}]=\pu{9.73E-4} \implies \text{pH} = -\log{h} = 3.01$$
Comparatively, the initial $[\ce{H+}] \approx \sqrt{K_\mathrm{a}.[\ce{NH4+}]}=\sqrt{\pu{5.2e-8}\cdot 0.1} \approx \pu{7.21E-5 mol L-1} \\ \implies \mathrm{pH} \approx 4.14$