= Solution
Write $V=\{1,\ldots,p\}$. The <global Markov property for a directed acyclic graph> says that, for every three pairwise disjoint vertex sets $A,B,C$,
$$
A\text{ and }B\text{ are D-separated by }C\text{ in }\mathcal G\quad\Longrightarrow\quad Z_A\mathrel{\perp\!\!\!\perp}Z_B\mid Z_C.
$$
Thus graphical <D-separation> implies probabilistic <conditional independence>. <Causal minimality> means that $P$ has this <global Markov property for a directed acyclic graph>, but has it for no proper subgraph of $\mathcal G$ obtained by deleting edges while keeping all vertices. <Faithfulness of a directed acyclic graph> means that the displayed implication is an equivalence: the <conditional independences> of $P$ are exactly the <D-separations> of $\mathcal G$. In particular it includes the <global Markov property for a directed acyclic graph>.
\b[A causally minimal distribution can nevertheless have an extra independence.] Here is a <causally minimal nonfaithful Gaussian model>. Take three <independent random variables> $E_1,E_2,E_3$, each with the <standard normal distribution>, and set
$$
Z_1=E_1,\qquad Z_2=Z_1+E_2,\qquad Z_3=Z_2-Z_1+E_3.
$$
Use the <Directed acyclic graph> with edges $1\to2$, $1\to3$ and $2\to3$. The equations give a <directed acyclic graph factorization>, so the <global Markov property for a directed acyclic graph> holds. Alternatively this complete acyclic graph imposes no nontrivial <D-separations>. But $Z_3=E_2+E_3$ is <independent> of $Z_1=E_1$, even though the direct edge $1\to3$ prevents their <D-separation>. Hence <faithfulness of a directed acyclic graph> fails.
To check <causal minimality> explicitly, the <covariance matrix> is
$$
\Sigma=\begin{pmatrix}1&1&0\\1&2&1\\0&1&2\end{pmatrix}.
$$
The linear transformation from $(E_1,E_2,E_3)$ to $(Z_1,Z_2,Z_3)$ is invertible, so this is a nonsingular <multivariate normal distribution>. Deleting $1\to2$ would require $Z_1\perp Z_2$, whereas $\operatorname{Cov}(Z_1,Z_2)=1$. Deleting $1\to3$ would require $Z_1\perp Z_3\mid Z_2$, whereas the conditional <covariance> is
$$
\operatorname{Cov}(Z_1,Z_3\mid Z_2)=0-\frac{1\cdot1}{2}=-\frac12.
$$
Deleting $2\to3$ would require $Z_2\perp Z_3\mid Z_1$, whereas its conditional <covariance> is $1-1\cdot0=1$. By <Gaussian conditional independence>, each nonzero conditional <covariance> rules out the required independence. Every proper subgraph is contained in a graph obtained by a single edge deletion, and further deletion can only add Markov restrictions. Thus no proper subgraph is Markov for $P$, proving <causal minimality>.
For the <Markov blanket>, use its standard convention that the focal vertex is excluded:
$$
B=\operatorname{mb}(k)=\bigl(\operatorname{pa}(k)\cup\operatorname{ch}(k)\cup\operatorname{pa}(\operatorname{ch}(k))\bigr)\setminus\{k\}.
$$
We prove <Markov blanket D-separation> by examining an arbitrary path from $k$ to $v\notin B\cup\{k\}$. If its first edge points into $k$, its next vertex is a parent of $k$, hence a conditioned noncollider in $B$, and the path is blocked. Otherwise the path starts $k\to h$ at a child $h\in B$. If the next edge points out of $h$, then $h$ is again a conditioned noncollider. If it points into $h$, the path begins $k\to h\leftarrow u$. Although conditioning on the <collider> $h$ opens that local segment, $u$ is another parent of $h$, hence belongs to $B$. The outgoing arrow $u\to h$ makes $u$ a noncollider on the path, so conditioning on $u$ blocks it. Since $v$ is outside $B$, none of these blocking vertices can be the endpoint.
Every path is therefore blocked by $B$. Applying the <global Markov property for a directed acyclic graph> gives
$$
\boxed{Z_k\mathrel{\perp\!\!\!\perp}Z_{V\setminus(B\cup\{k\})}\mid Z_B.}
$$
This proves the requested screening property without any positivity assumption on $P$.
Finally assume <faithfulness of a directed acyclic graph> and let $A$ have the stated screening property. Then $k$ and every vertex outside $A\cup\{k\}$ must be <D-separated> by $A$. If a parent or child $h$ of $k$ were missing from $A$, the one-edge path between them would be active, a contradiction. Thus every parent and child is in $A$. If another parent $u$ of a child $h$ were missing from $A$, the path $k\to h\leftarrow u$ would be active: its only interior vertex is the <collider> $h$, already in $A$. This is again a contradiction. Consequently
$$
\boxed{A\supseteq\operatorname{mb}(k).}
$$
\b[Under faithfulness the graphical blanket is contained in every screening set.] This is the <minimal Markov blanket under faithfulness>; the preceding argument also shows that the blanket itself is a screening set.
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