Let's take a look at the part of the diffusion operator between the columns of Hadamard gates. This part is supposed to perform a conditional phase shift, giving a phase of $-1$ to the state $|0...0\rangle$ and leaving the rest of the basis states unmodified.
For the first circuit, the bottom 3 wires are controls, wrapped in NOT gates, i.e., they are anti-controls: the operator applied on the 4th wire from the bottom only if each of the bottom 3 wires is in the $|0\rangle$ state. The operator performed on the 4th wire in this case is described by this circuit (two of the three NOT gates in the middle cancel right away):

Here are the transformations done by this circuit to a qubit in the starting state $\alpha |0\rangle + \beta |1\rangle$:
$$\alpha |0\rangle + \beta |1\rangle \xrightarrow{\oplus}
\beta |0\rangle + \alpha |1\rangle \xrightarrow{\text{H}}
\beta |+\rangle + \alpha |-\rangle \xrightarrow{\oplus}
\beta |+\rangle - \alpha |-\rangle \xrightarrow{\text{H}}
\beta |0\rangle - \alpha |1\rangle \xrightarrow{\oplus}
-\alpha |0\rangle + \beta |1\rangle$$
This is exactly what we're looking for - a $-1$ phase applied to the $|0...0\rangle$ state.
For the second circuit, the bottom 3 wires are already anti-controls, so the circuit we need to apply to the top two wires is the following:

The transformation is the following:
$$(\alpha |0\rangle + \beta |1\rangle) \otimes |-\rangle =
\frac{1}{\sqrt2} (\alpha |0\rangle |0\rangle - \alpha |0\rangle |1\rangle + \beta |1\rangle |0\rangle - \beta |1\rangle |1\rangle \xrightarrow{\text{CNOT}_0} $$
$$\frac{1}{\sqrt2} (\alpha |0\rangle \color{blue}{|1\rangle} - \alpha |0\rangle \color{blue}{|0\rangle} + \beta |1\rangle |0\rangle - \beta |1\rangle |1\rangle =
(\color{blue}{-} \alpha |0\rangle + \beta |1\rangle) \otimes |-\rangle$$
This transformation applies the same conditional phase shift to the bottom wire and does not modify the $|-\rangle$ state of the top wire, so it turns out to be equivalent to the first circuit.