= Solution
There is a normalization error in the original PDF: the first term contains $I$ rather than $\rho$. For a normalized <qubit> <density operator>, that printed expression has <trace> $2(1-p)+p=2-p$, so it is not a <quantum channel> except at $p=1$. The intended <depolarizing channel> is the <Pauli channel> that leaves the input unchanged with <probability> $1-p$ and applies each nonidentity <Pauli matrix> with <probability> $p/3$, for $0\leq p\leq1$. Its <Kraus operators> are
$$
K_0=\sqrt{1-p}\,I,\qquad K_k=\sqrt{p/3}\,\sigma_k\quad(k=1,2,3),
$$
and $\sum_{k=0}^3K_k^\dagger K_k=I$, proving that it is trace-preserving and a <completely positive map>.
Write the input in its <Bloch vector> representation,
$$
\rho=\frac12(I+\mathbf r\cdot\boldsymbol\sigma),\qquad |\mathbf r|\leq1.
$$
Conjugation by a <Pauli matrix> keeps its own <Bloch vector> component and reverses the other two. More explicitly, $\sigma_k\sigma_j\sigma_k=\sigma_j$ for $j=k$ and $-\sigma_j$ for $j\ne k$. Hence
$$
\sum_{k=1}^3\sigma_k\rho\sigma_k
=\frac12(3I-\mathbf r\cdot\boldsymbol\sigma)
=2I-\rho.
$$
The corrected <quantum channel> therefore gives
$$
\Phi_p(\rho)=(1-p)\rho+\frac p3(2I-\rho)
=\frac12\left[I+\left(1-\frac{4p}{3}\right)\mathbf r\cdot\boldsymbol\sigma\right].
$$
Thus its action is
$$
\boxed{\mathbf r\longmapsto\left(1-\frac{4p}{3}\right)\mathbf r.}
$$
The <Bloch sphere> is sent to a sphere of radius $|1-4p/3|$ about the origin of the <Bloch ball>. At $p=0$ the action is the identity. For $0<p<3/4$ the <Bloch vector> shrinks without changing direction. At $p=3/4$ every input becomes the maximally mixed <density operator> $I/2$. For $3/4<p\leq1$, the <Bloch vector> reverses direction and shrinks, reaching factor $-1/3$ at $p=1$. The reversal is compatible with complete positivity because this remains a probabilistic mixture of <Pauli matrices>. The parameter here is the total nonidentity-error <probability>, not the retention parameter used in some definitions of the <quantum depolarizing channel>.
For completeness, retain the PDF literally and call its output $F_p(\rho)$. The same calculation gives
$$
F_p(\rho)=\left(1-\frac p2\right)I-\frac p6\mathbf r\cdot\boldsymbol\sigma,
\qquad\operatorname{Tr}F_p(\rho)=2-p.
$$
This has no normalized-output <Bloch ball> interpretation as printed. If it is manually normalized, then
$$
\frac{F_p(\rho)}{2-p}
=\frac12\left[I-\frac{p}{3(2-p)}\mathbf r\cdot\boldsymbol\sigma\right].
$$
That different transformation sends $p=0$ to $I/2$, rather than to the input, and agrees with the intended answer only at $p=1$. The literal and corrected conventions therefore cannot be silently identified.
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