= Solution
Write $\lambda=(a,b)$ with $a\geq b>0$. If the removed corner is in the first row, it must be its last cell, with $a>b$. Remove that entry and output it as $v$; all remaining row and column inequalities persist, and the shape is $(a-1,b)$.
If the corner is the last cell of the second row, remove its value $c$. In the first row find the rightmost entry $d<c$, replace it by $c$, and output $v=d$. Such an entry exists: the first-row cell above the removed corner has smaller value than $c$. The chosen position is therefore at least column $b$.
The new first-row entry exceeds all entries to its left, while every entry to its right is larger than $c$ by the rightmost-choice rule. After deleting the lower corner, no second-row cell remains below the changed position, since that row now has <length> $b-1$. Thus all column inequalities remain valid too. The resulting shape is $(a,b-1)$, omitting a zero part if $b=1$.
$$
\boxed{\operatorname{shape}(P^-)=\lambda\setminus\{x\}.}
$$
The entries remain distinct and increasing along rows and columns, but the missing label is the ejected $v$, not necessarily the original corner value. The resulting tableau is therefore standard on its remaining ordered entry <set>, the generalized-standard convention in the question.
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