I guess you're referring to the Buckingham $\Pi$ Theorem statement from "Applied Mathematics", by J. David Logan:
Buckingham $\Pi$ Theorem
Let $f(q_1, \ldots, q_m)=0$ be a unit-free physical law that relates the dimensional quantities $q_1, \ldots, q_m$. Let $L_1, \ldots, L_n \; (n<m)$ be fundamental dimensions with $$\left[ q_i \right] = L^{a_{1i}}_{1} L^{a_{2i}}_{2} \cdots L^{a_{ni}}_{n}, \; i=1,\ldots,m,$$ and let $r = \text{rk}(D)$, where $D$ is the dimensionality matrix. Then there exist $m-r$ independent dimensionless quantities $\Pi_1, \Pi_2, \ldots, \Pi_{m-r}$ that can be formed from $q_1, \ldots, q_m,$ and the physical law $f(q_1, \ldots, q_m)=0$ is equivalent to an equation $$F \left( \Pi_1, \Pi_2, \ldots, \Pi_{m-r} \right)=0,$$ expressed only in terms of the dimensionless quantities.
The part you're struggling with is the $n < m$ bit, right?
I think that Logan states that condition in the theorem just as a reassurance that $\text{rk}(D) < m$, which is what you really need in order to be able to apply the theorem. (It probably makes his life easier when he is proving the theorem)
Now, if we assemble the dimensionality matrix of your specific problem, we get
$$D =
\begin{array}{c c} &
\begin{array}{c c c} E & P & A \\
\end{array}
\\
\begin{array}{c c c}
M \\
L \\
T
\end{array}
&
\left[
\begin{array}{r r r}
1 & 1 & 0 \\
2 & -1 & 2 \\
-2 & -2 & 0
\end{array}
\right]
\end{array}$$
and we can observe that the first and third rows of $D$ are linearly dependent, so $\text{rk}(D)=2$ and there is $m - \text{rk}(D) = 1$ dimensionless quantity, let's say $\Pi_1$.
Moreover,
$$ \text{Ker}(D) = \left\{ \alpha \begin{bmatrix}
-1 \\
1 \\
3/2 \\
\end{bmatrix}, \, \alpha \in \mathbb{R} \right\},$$
so, by choosing $\alpha=1$, we get
$$\Pi_1 = E^{-1} P^{1} A^{3/2}.$$
Hence, by the $\Pi$ Theorem, assuming that the physical law is unit-free, we conclude that
$$f(E,P,A) = 0,$$
is equivalent to the physical law
$$g(\Pi_1) = g\left(E^{-1} P^{1} A^{3/2}\right)=0,$$
and this implies that
$$E^{-1} P^{1} A^{3/2} = \text{const}.$$
