As Joel David Hamkins commented, this is likely to be a cardinal characteristic. Let me define (I hope temporarily) the relevant characteristic $\mathfrak{dfpmk}$ (named after the OP and the author of the accepted answer) to be the smallest cardinal $\kappa$ such that, for some non-principal ultrafilters $\mathcal U$ and $\mathcal V$ on $\omega$ and some family of $\kappa$ sequences $(s_n^\xi)_{n\in\omega}$ (indexed by $\xi\in\kappa$), the sequences all converge to limits $\lambda^\xi$ with respect to $\mathcal U$ but no simultaneous rearrangements $(s_{r(n)}^\xi)_{n\in\omega}$ all converge to the corresponding $\lambda^\xi$ with respect to $\mathcal V$. So what Paul McKenney proved is that $\mathfrak p\leq\mathfrak{dfpmk}\leq\mathfrak u$.

Of course, what Joel probably intended is that $\mathfrak{dfpmk}$ might be a *known* cardinal characteristic. I can't prove anything like that, but I can improve in *some* models the upper bound $\mathfrak u$ in Paul's answer. Namely, if there exists a P-point, then $\mathfrak{dfpmk}\leq\mathfrak d$. (This is an improvement only if there is a P-point and $\mathfrak d<\mathfrak u$; there are models where that happens, for example, the random real model.)

Before giving the proof, let me indicate an alternative way to view $\mathfrak{dfpmk}$, which will be useful (at least for me) in the proof. To avoid excess verbiage, assume that all filters in the following extend the cofinite filter on $\omega$, so in particular all ultrafilters are non-principal.) I claim that $\mathfrak{dfpmk}$ is the smallest number of generators needed for a filter $\mathcal F$ on $\omega$ such that, for some ultrafilter $\mathcal U$, no isomorphic copy of $\mathcal U$ extends $\mathcal F$.

To prove the equivalence, assume first that $\kappa<\mathfrak{dfpmk}$, let $\mathcal F$ be any filter with a basis $\mathcal B$ of size $\kappa$, and let $\mathcal U$ be any ultrafilter. Consider the characteristic functions of the sets in $\mathcal B$ and consider any ultrafilter $\mathcal V$ extending $\mathcal F$. These characteristic functions all converge to 1 with respect to $\mathcal V$, and there are only $\kappa<\mathfrak{dfpmk}$ of them, so some simultaneous rearrangements of them, say by a permutation $r$, all converge to 1 with respect to $\mathcal U$. But that means that the original (not rearranged) characteristic functions converge to 1 with respect to $r^{-1}(\mathcal U)$. That means $r^{-1}(\mathcal U)$ contains all the sets in $\mathcal B$ and therefore extends $\mathcal F$.

For the converse, suppose $\kappa$ is smaller than my proposed alternative view of $\mathfrak{dfpmk}$, and let $\kappa$ sequences $(s_n^\xi)$ converge to limits $\lambda^\xi$ with respect to an ultrafilter $\mathcal U$.This convergence means that $\mathcal U$ contains all the sets $A_{\xi,k}=\{n\in\omega: |s_n^\xi-\lambda^\xi|<2^{-k}\}$ for $\xi<\kappa$ and $k\in\omega$. These $\kappa$ sets $A_{\xi,k}$ generate a filter $\mathcal F$, and, by choice of $\kappa$, every ultrafilter $\mathcal V$ has an isomorphic copy $r(\mathcal V)$ extending $\mathcal F$. Then the sequences $(s_n^\xi)$ converge to $\lambda^\xi$ with respect to $r(\mathcal V)$, and therefore the simultaneous rearrangements $(s_{r^{-1}(n)}^\xi)$ converge to $\lambda^\xi$ with respect to $\mathcal V$.

This completes the proof of the equivalence between the two views of $\mathfrak{dfpmk}$. Now to prove my claim about $\mathfrak d$, consider the filter $\mathcal F$ on the "plane" $\omega\times\omega$ consisting of those sets $X$ such that, for all but finitely many $x\in\omega$, the "column" at abscissa $x$ has all but finitely many of its points in $X$. That is, for all but finitely many $x$, for all but finitely many $y$, $(x,y)\in X$. (This filter is often called the tensor square or the Fubini square of the cofinite filter on $\omega$.) This filter is generated by $\mathfrak d$ sets, namely the sets $\{(x,y):x>n\}$ for each $n\in\omega$ and the sets $\{(x,y):y>f(x)\}$ for each $f$ in some dominating family of cardinality $\mathfrak d$.

If an ultrafilter $\mathcal U$ extends $\mathcal F$, then $\mathcal U$ cannot be a P-point, because any set on which the projection to the first factor, $(x,y)\mapsto x$, is finite-to-one or constant is the complement of a set if $\mathcal F$ and therefore cannot be in $\mathcal U$. Furthermore, since the property of being a P-point is preserved by isomorphism, no P-point can be isomorphic to an extension of $\mathcal F$. So, as long as there is a P-point, the filter $\mathcal F$ witnesses, in the alternative view of $\mathfrak{dfpmk}$ above, that $\mathfrak{dfpmk}\leq\mathfrak d$.

A very similar argument shows, under the weaker hypothesis that there exists a nowhere dense ultrafilter, that $\mathfrak{dfpmk}\leq\mathfrak{cof}(B)$. (Here "nowhere dense" is in the sense of Baumgartner's $I$-ultrafilters; it means that the image of $\mathcal U$ under any map $\omega\to\mathbb Q$ contains a nowhere dense set. $\mathcal{cof}(B)$ is the cofinality number for Baire category, the minimum number of sets needed to generate the ideal of meager subsets of $\mathbb R$.)