It seems that Kraus operators cannot change a pure state into a mixed one (wrong). For any pure state can be written as $|\psi\rangle\langle\psi|$, so after the Kraus operators. It becomes $$\sum_l\Pi_l|\psi\rangle\langle\psi|\Pi_l^\dagger = |\phi\rangle\langle\phi|.$$
But does there exist some Kraus operators that can change the mixed state $\rho$ into a pure state?
More generally, given any two states, you can always find some channel sending one into the other. Consider for example replacement maps, which have the form $$\Phi_Y(X) = \operatorname{Tr}(X) Y.$$ Given any pair of states $\rho$ and $\sigma$, the channel $\Phi_\sigma$ will send $\rho$ (as well as any other state) into $\sigma$. The (or a) set of Kraus operators for $\Phi_Y$ is $\{\sqrt{y_\alpha}\lvert y_\alpha\rangle\!\langle \beta|\}_{\alpha,\beta}$, where $|y_\alpha\rangle$ form an orthonormal basis of eigenvectors for $Y$ (assuming $Y$ to be normal), $y_\alpha$ are the corresponding eigenvalues, and $|\beta\rangle$ is an arbitrary orthonormal basis (for the relevant space).
See also @Norbert's answer on physics.SE for other examples of quantum channels on which to test hypotheses.
To add to the already great answers let me point out that if the initial mixed state $\rho$ has full rank, then there is only one type of channel which can send that state to a pure state $|\phi\rangle\langle\phi|$: the replacement/reset channel $X\mapsto{\rm tr}(X)|\phi\rangle\langle\phi|$. The reason for this is that all positive maps $\Phi$ "increase kernels" in the sense that given any $X,Y\geq 0$ with ${\rm ker}(X)\subseteq{\rm ker}(Y)$ one always has ${\rm ker}(\Phi(X))\subseteq{\rm ker}(\Phi(Y))$ (equivalently: positive maps increase ranges) as is shown, e.g., in Lemma 1.1 of this paper or Proposition 3.2 of one of my old papers.
Either way this kernel property implies that if a channel $\Phi$ maps $\rho>0$ to $|\phi\rangle\langle\phi|$, then for all states $\omega$---because $\{0\}={\rm ker}(\rho)\subseteq{\rm ker}(\omega)$---it holds that $|\phi\rangle^\perp={\rm ker}(\Phi(\rho))={\rm ker}(\Phi(\omega))$, i.e. $\Phi(\omega)=|\phi\rangle\langle\phi|$. By linearity this shows that $\Phi(X)={\rm tr}(X)|\phi\rangle$ for all $X$, as claimed. More generally, the above kernel property places a restriction on all channels that map some non-pure state to a pure state. A reference for this more special result is, e.g., Lemma 5.6 in this paper
$\left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right) $ and $\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right) $ can change one mixed state into a pure state.
