Describing the crystalline extension of $\mathbb{Q}_p$ by $\mathbb{Q}_p$

The representation $E$ in this case is not only crystalline, it is in fact unramified. This means we don't need much of the complicated machinery of $p$-adic Hodge theory to get a handle on the periods of $E$.

Whereas for general crystalline representations we need to use $\mathbf{B}_\text{cris}$ to find periods, for potentially unramified representations, we can work with $\mathbb{C}_p$ (potentially unramified representations are $\mathbb{C}_p$-admissible). In this case, we don't even need $\mathbb{C}_p$, as the representation is unramified (not just potentially so), and it suffices to work with $(K_0^\text{nr})^\vee$ instead (where $K_0$ is the maximal unramified intermediate extension $K / K_0 / \mathbb{Q}_p$).

To explicitly see that $E$ is $(K_0^\text{nr})^\vee$-admissible, we can start by taking $a$ to be a solution to the Artin–Schreier equation $x^q-x-1=0$ in $\mathcal{O}_K/\mathfrak{m} \cong \mathbb{F}_q$. We then have $\mathrm{Frob}([a]) = [a^q] = [a+1]$, where $[-]$ indicates Teichmüller lifts. This is nearly what we want, save for the fact that the Teichmüller lift is not additive. So you have to remedy that by hand by using the Witt addition polynomials; the upshot is that you'll obtain some element $b \in \mathrm{W}(\overline{\mathbb{F}_q})$ with $\varphi(b) = b+1$ as desired. It is necessary to go all the way up to $\mathrm{W}(\overline{\mathbb{F}_q})$; at any finite level $\mathrm{W}(\mathbb{F}_{q^n})$ there will always be the above issue of non-additivity. Indeed, only potentially trivial representations are detected at finite levels, and we need to pass to the completion $(K_0^\text{nr})^\vee$ to allow a Frobenius of infinite order.
At any rate, this means then that $\{1,b\}$ is a $(K_0^\text{nr})^\vee$-basis of $E \otimes_{\mathbb{Q}_p} (K_0^\text{nr})^\vee$. Using the inclusion $(\mathcal{O}_{K_0^\text{nr}})^\vee \subset \mathbf{A}_\text{cris}$ we can consider $b$ to be an element of $\mathbf{B}_\text{cris}$. It is a crystalline period of $E$; together with $1 \in \mathbf{B}_\text{cris}$ it provides a $\textbf{B}_\text{cris}$-basis of $E \otimes_{\mathbb{Q}_p} \mathbf{B}_\text{cris}$.

As for the motivic question, I think no such variety is expected to exist. We are in the crystalline situation, so we'd like to be able to find $E$ inside the cohomology of some (smooth, separated, finite type) scheme over $\mathcal{O}_K/\mathfrak{m}$. By analogy with the $\ell$-adic and complex situations, I believe such extensions as $E$ shouldn't occur there, because of weight filtration considerations (e.g. there are no non-split extensions of mixed Hodge structures of $\mathbb{Q}$ by $\mathbb{Q}$).

For simplicity, let's take $K = \mathbf{Q}_p$.

One of the few things about $\mathbf{B}_{\mathrm{cris}}$ that one can straightforwardly prove directly from its definition is that it contains $\widehat{\mathbf{Q}}_p^{\mathrm{nr}}$, the completion of the maximal unramified extension of $\mathbf{Q}_p$ (equvalently, the field of fractions of the Witt vectors of $\overline{\mathbf{F}}_p$).

It's not too hard to check, using Hilbert 90 for $\overline{\mathbf{F}}_p$ and induction on $n$, that any unramified mod $p^n$ representation of $G_{\mathbf{Q}_p}$ is $W_n(\overline{\mathbf{F}}_p)$-admissible, and by passage to the limit ("devissage", as the French call it) one deduces that any unramified $\mathbf{Q}_p$-linear representation of $G_{\mathbf{Q}_p}$ is $\widehat{\mathbf{Q}}_p^{\mathrm{nr}}$-admissible.

From local class field theory, one knows that there's a unique unramfied extension of $\mathbf{Q}_p$ by $\mathbf{Q}_p$ (corresponding to the unramfied $\mathbf{Z}_p$-extension of $\mathbf{Q}_p$). So this representation is crystalline, and its periods lie in this much less scary subring $\widehat{\mathbf{Q}}_p^{\mathrm{nr}} \subset \mathbf{B}_{\mathrm{cris}}$.

I'm not so sure whether there's a nice way of exhibiting this extension in the etale cohomology of a variety. The variety would have to be either non-smooth or non-proper (since the cohomology of smooth proper varieties is conjectured to be semi-simple). Maybe someone with more of a geometric background than me can answer that.