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Compilation-supplied correction. Welsh's prose in Theorem 9, printed p. 1327 (published PDF), asks this question, but its displayed criterion is false at that scope.

Statement. Let $U\subseteq S$ and let $N=p(I)$. There is a $p$-transversal $X$ of $\mathcal A$ with $U\subseteq X$ if and only if, for every $J\subseteq I$,

$$|A(J)|\ge p(J), \tag{1}$$

and

$$|A(J)\cap U|\ge p(J)+|U|-N. \tag{2}$$

Proof. Suppose $X=\bigsqcup_{i\in I}X_i$ is a $p$-transversal containing $U$. The union of the $J$-parts is a $p(J)$-element subset of $A(J)$, proving (1). At most $N-p(J)$ elements of $X$, and hence at most that many elements of $U$, lie in parts indexed outside $J$. Therefore

$$|U\setminus A(J)|\le N-p(J),$$

which rearranges to (2).

Conversely, assume (1) and (2). Condition (1) and Theorem 7 show that the replicated family $\mathcal A^p$ has a full transversal.

On $S$, take the matroid in which the elements of $U$ are free and all elements of $S\setminus U$ are loops. Its rank is

$$r_U(Y)=|Y\cap U|.$$

For a subfamily $K\subseteq I^p$ of copied indices, let $J$ be its support in $I$. Then $|K|\le p(J)$ and its union is $A(J)$. Condition (2) gives

$$r_U(A(J)) =|A(J)\cap U| \ge |U|+p(J)-N \ge |U|+|K|-N.$$

By Perfect's criterion, $\mathcal A^p$ has a partial-transversal range $P$ with $r_U(P)\ge|U|$. Since $r_U(P)\le|U|$, this says $U\subseteq P$. The copied family also has a full transversal, so augmentation extends $P$ to a full-transversal range $X$. Then $U\subseteq X$, and the replicated-family correspondence makes $X$ a $p$-transversal of $\mathcal A$.

At $J=\varnothing$, (2) forces $|U|\le N$. When $U=\varnothing$, (2) is automatic and the statement reduces to the ordinary $p$-Hall criterion. $\square$

The use of Perfect's criterion keeps the exact finite Rado input described in external inputs.