= Solution
The PDF's first Pauli term contains an undefined $\rho_x$. For the stated <depolarizing channel>, use $\sigma_x\rho\sigma_x$, consistently with the other two terms. This is the minimal typographical correction needed to define the intended <Pauli channel>.
Write $\rho=\frac12(I+\mathbf s\cdot\boldsymbol\sigma)$ using its <Bloch vector>. The <Pauli matrices> satisfy $\sigma_j\sigma_k\sigma_j=\sigma_k$ for $j=k$ and $-\sigma_k$ for $j\ne k$. Thus conjugating by each nonidentity Pauli matrix leaves one Bloch component unchanged and reverses the other two. Summing the three conjugates gives
$$
\sum_{j=x,y,z}\sigma_j\rho\sigma_j=\frac12(3I-\mathbf s\cdot\boldsymbol\sigma)=2I-\rho.
$$
Substitution yields
$$
\Phi(\rho)=\left(1-\frac{4p}{3}\right)\rho+\frac{2p}{3}I,
$$
so the <Pauli-mixture parametrization of qubit depolarization>, with $p$ here the total error probability, gives
$$
\boxed{q=\frac{4p}{3},\qquad\Phi(\rho)=(1-q)\rho+qI/2.}
$$
There is also a genuine range error in the printed claim. For its full $0<p<1$ assumption, $0<q<4/3$, and \b[$0<q<1$ holds exactly when $0<p<3/4$]. For example $p=9/10$ gives $q=6/5$; on $|0\rangle\langle0|$ the output is $\operatorname{diag}(2/5,3/5)$, with negative $z$ Bloch component. No $q\in(0,1)$ could produce this output from that input, since its component would be $1-q>0$.
The corrected channel is still completely positive in the larger range: its original Pauli probabilities are $1-p,p/3,p/3,p/3$. This is <depolarizing noise weight above one>; the formula with $q>1$ is not a convex mixture with weights $1-q,q$, but it has a valid <random unitary channel> representation. The subsequent parts can therefore be solved both in the printed $q\in(0,1)$ regime and in the actual parameter range.
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