Let $\Lambda$ be a ring, $V$ a $\Lambda$-algebra complete Hausdorff for the $p$-adic topology where $p$ a a prime integer. Assume that $V/pV$ is semiperfect (i.e., the Frobenius map is surjective). Let $R_V$ (also denoted $V^{\flat}$) the tilt of $V$, that is, the projective limit of $...\to V/pV\to V/pV$ where the maps are Frobenius. The elements $x$ of $R_V$ are identified to sequences $\{x^{(m)}\}_{m\geq 0}$ in $V$ such that $(x^{(m+1)})^p=x^{(m)}$ for all $m$ (the reduction mod $p$ of this sequence gives the sequence defining $x$ as element of the projective limit). Let $W(R_V)$ be the ring of Witt vectors and $$W(R_V)\to V$$ the ring homomorphism sending $[x_0,x_1,x_2,...]$ to $\sum x_n^{(n)}p^n$, which defines a homomorphism of $\Lambda$-algebras $$\theta :\Lambda\otimes_{\mathbb Z}W(R_V)\to V$$.
In "Le corps des périodes p-adiques", Fontaine defines $${\rm A_{inf}}(V|\Lambda)$$ as the completion of $\Lambda\otimes_{\mathbb Z}W(R_V)$ for the $(p)+{\rm ker}(\theta)$-adic topology, and in remark 1.2.4.(b) he says that the canonical map $${\rm A_{inf}}(V|\Lambda)\to {\rm A_{inf}}(V/pV|\Lambda)$$ is an isomorphism.
Why?
It is obvious that $R_V=R_{V/pV}$, but I am not able to prove that the completions are isomorphic.
Calling $\overline V:= V/pV$, there is a commutative diagram
$\require{AMScd}$ \begin{CD} \Lambda \otimes W(R_V) @>{\theta_1}>> V\\ @V\simeq ViV @VVV\\ \Lambda \otimes W(R_{\overline{V}}) @>{\theta_2}>> \overline V \end{CD}
where the left vertical arrow is the isomorphism $i$ induced from the natural isomorphism $R_V \simeq R_{\overline{V}}$ you have, the right vertical arrow is the natural mod $p$ projection, and the horizontal maps are the $\theta$'s for constructing $A_{inf}(V\vert \Lambda)$ resp. $A_{inf}(\overline{V}\vert \Lambda)$.
From here, $i(ker(\theta_1) + (p)) = ker(\theta_2)+(p)$ and the claim is immediate.