This is true with $\mathbb Q$-coefficients; see the proposition below. The reason it's difficult is that it's hard to compute Chow groups over non-algebraically closed fields. However, we have the following:
>> **Lemma 1.** *Let $k \subseteq \ell$ be a separable algebraic field extension, and let $X$ be a proper geometrically integral $k$-scheme. Then the pullback map $\operatorname{Pic}(X) \to \operatorname{Pic}(X_\ell)$ is injective.*
*Proof.* See e.g. [this post][1]. $\square$
>> **Lemma 2.** *Let $k \subseteq \ell$ be a finite field extension, and let $X$ be any $k$-scheme. Then the pullback map $\operatorname{CH}_*(X) \otimes_\mathbb Z \mathbb Q \to \operatorname{CH}_*(X_\ell) \otimes_\mathbb Z \mathbb Q$ is injective.*
*Proof.* If $\pi \colon X_\ell \to X$ is the map induced by $k \to \ell$, then $\pi_* \pi^*$ is multiplication by $[\ell:k]$; see e.g. [Ful, Ex. 1.7.4]. (This is in fact true for *any* finite flat morphisms of schemes.) $\square$
However, there exist Severi-Brauer varieties $X$ for which the pullback $\operatorname{CH}_*(X) \to \operatorname{CH}_*(X_\ell)$ (with integral coefficients) is *not* injective, as is explained in [this post][2].
Let $C$ be a smooth conic without rational points, and let $k \to \ell$ be the minimal splitting field (which is a separable quadratic extension of $k$).
>> **Lemma 3.** *We have $\operatorname{CH}^*(C) = \mathbb Z[2h] \subseteq \mathbb Z[h]/(h^2) = \operatorname{CH}^*(C_\ell)$.*
*Proof.* The pullback $\operatorname{CH}^*(C) \to \operatorname{CH}^*(C_\ell)$ is injective by **Lemma 1** above. The line bundle $\omega_C^{-1}$ has class $2h$. If $\mathscr L$ is a line bundle with class $h$, then Riemann-Roch gives an isomorphism $C \stackrel\sim\to \mathbb P^1$, contradicting the choice of $C$. $\square$
>> **Lemma 4.** *The image of the pullback $\operatorname{CH}^*(C \times C) \to \operatorname{CH}^*((C\times C)_\ell)$ equals*
$$\mathbb Z[h_1+h_2,2h_1] \subseteq \mathbb Z[h_1,h_2]/(h_1^2,h_2^2).$$
*Proof.* Clearly $h_1 + h_2 = [\Gamma_{\operatorname{id}_C}]$ and $2h_1 = \omega_C^{-1} \boxtimes \mathcal O_C$ are contained in $\operatorname{CH}^*(C \times C)$. Moreover, they span $\operatorname{CH}^1(C \times C) = \operatorname{Pic}(C \times C)$, since $h_1$ is not defined on $C \times C$. Indeed, if $\mathscr L$ is a line bundle with class $h_1$, then it defines a map $C \times C \to \mathbb P^1$, whose image under the $2$-uple embedding is the first projection $C \times C \to C \subseteq \mathbb P^2$. This again contradicts the choice of $C$.
Finally, the intersection $(h_1 + h_2) \cdot 2h_1 = 2h_1h_2$ is a degree $2$ zero-cycle on $C \times C$. There are no zero-cycles of odd degree, for their pushforward under either projection would be a divisor of odd degree on $C$. $\square$
**Remark.** I don't know if the pullback $\operatorname{CH}_*(C \times C) \to \operatorname{CH}^*((C \times C)_\ell)$ is *injective*. This is true for divisors (and for dimension $2$ cycles), and it is true rationally. However, there could be torsion zero-cycles that disappear on $(C \times C)_\ell$.
**Remark.** We will use the ring isomorphism $\operatorname{CH}^*(X \times \mathbb P^1) \cong \operatorname{CH}^*(X) \otimes \mathbb Z[h]/h^2$ of [Ful, Ex. 8.3.4] for any smooth $k$-variety $X$ (it is true as groups in much greater generality). Then the pushfoward $\operatorname{CH}^*(X \times \mathbb P^1) \to \operatorname{CH}^*(X)$ is given by $\alpha \otimes 1 \mapsto 0$ and $\alpha \otimes h \mapsto \alpha$ for $\alpha \in \operatorname{CH}^*(X)$.
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>> **Proposition.** *The motives $C$ and $\mathbb P^1$ are isomorphic rationally.*
*Proof.* With rational coefficients, we have $\operatorname{CH}_\mathbb Q^*(C) \cong \operatorname{CH}_\mathbb Q^*(\mathbb P^1) \cong \mathbb Q[h]/(h^2)$, and similarly for all finite products involving $C$ and $\mathbb P^1$. The class $\phi = \tfrac{1}{2}[\omega_C^{-1} \boxtimes \mathcal O_{\mathbb P^1}(2)] \in \operatorname{Corr}(C, \mathbb P^1)$ corresponds to $h_1+h_2$. We now easily compute
\begin{align*}
\phi \circ \phi^\top &= \pi_{13,*}(\pi_{12}^* \phi^\top \cdot \pi_{23}^* \phi) = \pi_{13,*}((h_1+h_2)(h_2+h_3)) \\
&= \pi_{13,*}(h_1h_3 + h_2(h_1 + h_3) ) = h_1 + h_3 \\
&= [\Gamma_{\operatorname{id}_{\mathbb P^1}}] \in \operatorname{CH}^*_\mathbb Q(\mathbb P^1 \times \mathbb P^1),
\end{align*}
and similarly for the other composition $\phi^\top \circ \phi$. $\square$
**Remark.** I think that the motive $C$ does not occur integrally as a direct summand of $\mathbb P^n \times Z$ for any $n \in \mathbb Z_{\geq 0}$ and any zero-dimensional smooth $k$-scheme $Z$.
For example, if $Z = \operatorname{Spec} m$ for a Galois extension $k \subseteq m$ with group $G$, then $H^0(C_\bar k, \mathbb F_p)$ should occur as a direct summand of
$$H^0((\operatorname{Spec}_\ell)_\bar k, \mathbb F_p) = \mathbb F_p[G]$$
for any prime $p \not\mid \operatorname{char} k$.
But if $p \mid \#G$, then the unique trivial subrepresentation $\mathbb F_p \cdot (\sum_{\sigma \in G} \sigma) \subseteq \mathbb F_p[G]$ does not split as a direct summand. If $\operatorname{char} k = 0$, this forces $G = 1$, i.e. $Z = \operatorname{Spec} k$. Then it remains to compute $\operatorname{CH}^*(C \times \mathbb P^n)$ as above and see that no $\phi$ in there satisfies $\phi \circ \phi^\top = 1$, even modulo projectors.
This is not a very good argument, so I would be interested to see if someone has a cleaner obstruction for the integral version of the question.
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**References.**
[Ful] <cite authors="Fulton, William"> W. Fulton, *Intersection theory*, second edition. Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. **2**. Springer-Verlag (1998). [ZBL0885.14002](https://zbmath.org/?q=an:0885.14002).</cite>
[1]: https://mathoverflow.net/a/180188/82179
[2]: https://mathoverflow.net/a/289803/82179