We will prove this using Lemma \ref{lemma-find-homotopy} with $a = p$. Thus we have to find $\theta : D' \to D'$ and prove (\ref{item-d-a-zero}), (\ref{item-direct-sum}), (\ref{item-theta-linear}), (\ref{item-integrate}), (\ref{item-injective}), (\ref{item-factor}), (\ref{item-horizontal}). We first collect some information about the rings $D$ and $D'$ and the modules $\Omega_D$ and $\Omega_{D'}$. \medskip\noindent Writing $$ D[z]\langle \xi \rangle/(\xi - (z^p - \lambda)) = D\langle \xi \rangle[z]/(z^p - \xi - \lambda) $$ we see that $D'$ is the $p$-adic completion of the free $D$-module $$ \bigoplus\nolimits_{i = 0, \ldots, p - 1} \bigoplus\nolimits_{n \geq 0} z^i \xi^{[n]} D $$ where $\xi^{[0]} = 1$. It follows that $D \to D'$ has a continuous $D$-linear section, in particular $D \to D'$ is universally injective, i.e., (\ref{item-injective}) holds. We think of $D'$ as a divided power algebra over $A$ with divided power ideal $\overline{J}' = \overline{J}D' + (\xi)$. Then $D'$ is also the $p$-adic completion of the divided power envelope of the ideal generated by $z^p - \lambda$ in $D$, see Lemma \ref{lemma-describe-divided-power-envelope}. Hence $$ \Omega_{D'} = \Omega_D \otimes_D^\wedge D' \oplus D'\text{d}z $$ by Lemma \ref{lemma-module-differentials-divided-power-envelope}. This proves (\ref{item-direct-sum}). Note that (\ref{item-d-a-zero}) is obvious. \medskip\noindent At this point we construct $\theta$. (We wrote a PARI/gp script theta.gp verifying some of the formulas in this proof which can be found in the scripts subdirectory of the Stacks project.) Before we do so we compute the derivative of the elements $z^i \xi^{[n]}$. We have $\text{d}z^i = i z^{i - 1} \text{d}z$. For $n \geq 1$ we have $$ \text{d}\xi^{[n]} = \xi^{[n - 1]} \text{d}\xi = - \xi^{[n - 1]}\text{d}\lambda + p z^{p - 1} \xi^{[n - 1]}\text{d}z $$ because $\xi = z^p - \lambda$. For $0 < i < p$ and $n \geq 1$ we have \begin{align*} \text{d}(z^i\xi^{[n]}) & = iz^{i - 1}\xi^{[n]}\text{d}z + z^i\xi^{[n - 1]}\text{d}\xi \\ & = iz^{i - 1}\xi^{[n]}\text{d}z + z^i\xi^{[n - 1]}\text{d}(z^p - \lambda) \\ & = - z^i\xi^{[n - 1]}\text{d}\lambda + (iz^{i - 1}\xi^{[n]} + pz^{i + p - 1}\xi^{[n - 1]})\text{d}z \\ & = - z^i\xi^{[n - 1]}\text{d}\lambda + (iz^{i - 1}\xi^{[n]} + pz^{i - 1}(\xi + \lambda)\xi^{[n - 1]})\text{d}z \\ & = - z^i\xi^{[n - 1]}\text{d}\lambda + ((i + pn)z^{i - 1}\xi^{[n]} + p\lambda z^{i - 1}\xi^{[n - 1]})\text{d}z \end{align*} the last equality because $\xi \xi^{[n - 1]} = n\xi^{[n]}$. Thus we see that \begin{align*} \partial_z(z^i) & = i z^{i - 1} \\ \partial_z(\xi^{[n]}) & = p z^{p - 1} \xi^{[n - 1]} \\ \partial_z(z^i\xi^{[n]}) & = (i + pn) z^{i - 1} \xi^{[n]} + p \lambda z^{i - 1}\xi^{[n - 1]} \end{align*} Motivated by these formulas we define $\theta$ by the rules $$ \begin{matrix} \theta(z^j) & = & p\frac{z^{j + 1}}{j + 1} & j = 0, \ldots p - 1, \\ \theta(z^{p - 1}\xi^{[m]}) & = & \xi^{[m + 1]} & m \geq 1, \\ \theta(z^j \xi^{[m]}) & = & \frac{p z^{j + 1} \xi^{[m]} - \theta(p\lambda z^j \xi^{[m - 1]})}{(j + 1 + pm)} & 0 \leq j < p - 1, m \geq 1 \end{matrix} $$ where in the last line we use induction on $m$ to define our choice of $\theta$. Working this out we get (for $0 \leq j < p - 1$ and $1 \leq m$) $$ \theta(z^j \xi^{[m]}) = \textstyle{\frac{p z^{j + 1} \xi^{[m]}}{(j + 1 + pm)} - \frac{p^2 \lambda z^{j + 1} \xi^{[m - 1]}}{(j + 1 + pm)(j + 1 + p(m - 1))} + \ldots + \frac{(-1)^m p^{m + 1} \lambda^m z^{j + 1}} {(j + 1 + pm) \ldots (j + 1)}} $$ although we will not use this expression below. It is clear that $\theta$ extends uniquely to a $p$-adically continuous $D$-linear map on $D'$. By construction we have (\ref{item-theta-linear}) and (\ref{item-integrate}). It remains to prove (\ref{item-factor}) and (\ref{item-horizontal}). \medskip\noindent Proof of (\ref{item-factor}) and (\ref{item-horizontal}). As $\theta$ is $D$-linear and continuous it suffices to prove that $p - \theta \circ \partial_z$, resp.\ $(\theta \otimes 1) \circ \text{d}_1 - \text{d}_1 \circ \theta$ gives an element of $D$, resp.\ $\Omega_D$ when evaluated on the elements $z^i\xi^{[n]}$\footnote{This can be done by direct computation: It turns out that $p - \theta \circ \partial_z$ evaluated on $z^i\xi^{[n]}$ gives zero except for $1$ which is mapped to $p$ and $\xi$ which is mapped to $-p\lambda$. It turns out that  $(\theta \otimes 1) \circ \text{d}_1 - \text{d}_1 \circ \theta$ evaluated on $z^i\xi^{[n]}$ gives zero except for $z^{p - 1}\xi$ which is mapped to $-\lambda$.}. Set $D_0 = \mathbf{Z}_{(p)}[\lambda]$ and $D_0' = \mathbf{Z}_{(p)}[z, \lambda]\langle \xi \rangle/(\xi - z^p + \lambda)$. Observe that each of the expressions above is an element of $D_0'$ or $\Omega_{D_0'}$. Hence it suffices to prove the result in the case of $D_0 \to D_0'$. Note that $D_0$ and $D_0'$ are torsion free rings and that $D_0 \otimes \mathbf{Q} = \mathbf{Q}[\lambda]$ and $D'_0 \otimes \mathbf{Q} = \mathbf{Q}[z, \lambda]$. Hence $D_0 \subset D'_0$ is the subring of elements annihilated by $\partial_z$ and (\ref{item-factor}) follows from (\ref{item-integrate}), see the discussion directly preceding Lemma \ref{lemma-find-homotopy}. Similarly, we have $\text{d}_1(f) = \partial_\lambda(f)\text{d}\lambda$ hence $$ \left((\theta \otimes 1) \circ \text{d}_1 - \text{d}_1 \circ \theta\right)(f) = \left(\theta(\partial_\lambda(f)) - \partial_\lambda(\theta(f))\right) \text{d}\lambda $$ Applying $\partial_z$ to the coefficient we obtain \begin{align*} \partial_z\left( \theta(\partial_\lambda(f)) - \partial_\lambda(\theta(f)) \right) & = p \partial_\lambda(f) - \partial_z(\partial_\lambda(\theta(f))) \\ & = p \partial_\lambda(f) - \partial_\lambda(\partial_z(\theta(f))) \\ & = p \partial_\lambda(f) - \partial_\lambda(p f) = 0 \end{align*} whence the coefficient does not depend on $z$ as desired. This finishes the proof of the lemma.