I was confused about this as well. The simplest way I can think of to do this is to use the \vphantom command within the table. Here are two tables -- one without the \vphantom and one with the \vphantom:
Without \vphantom
\begin{table}[htpb]
\centering
\pgfplotstabletypeset[
col sep=&,
row sep=\\,
string type,
columns = {x, N-x, pq, binom,prob},
columns/x/.style={column name={\boldmath{$x$}\unboldmath},
column type={@{}M{.05\dimexpr\textwidth-6\tabcolsep}}} ,
columns/N-x/.style={column name={\boldmath{$N-x$}\unboldmath},
column type={M{.1\dimexpr\textwidth-6\tabcolsep} }} ,
columns/pq/.style={column name={\boldmath{$p^{x}\ q^{N - x}$}\unboldmath},
column type={M{.3\dimexpr\textwidth-6\tabcolsep} }} ,
columns/binom/.style={column name={\boldmath{$\dfrac{N!}{x! (N - x)!}$}\unboldmath},
column type={M{.3\dimexpr\textwidth-6\tabcolsep}@{} }} ,
columns/prob/.style={column name={\boldmath{$P(x)$}\unboldmath},
column type={M{.1\dimexpr\textwidth-6\tabcolsep}@{} }} ,
every head row/.style=
{before row=\toprule\toprule, after row=\midrule},
every last row/.style=
{after row=\bottomrule\bottomrule},
]
{
x & N-x & pq & binom & prob \\
{}3 & 0 & $\left(\dfrac{1}{2}\right)^3\left(\dfrac{1}{2}\right)^0 = \dfrac{1}{8}$ & $\dfrac{3!}{3! (0!)} = 1$ & $\dfrac{1}{8}$ \\
\midrule
{}2 & 1 & $\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}\right)^1 = \dfrac{1}{8}$ & $\dfrac{3!}{2! (1!)} = 3$ & $\dfrac{3}{8}$ \\
\midrule
{}1 & 2 & $\left(\dfrac{1}{2}\right)^1\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{8}$ & $\dfrac{3!}{1! (2!)} = 3$ & $\dfrac{3}{8}$ \\
\midrule
{}0 & 3 & $\left(\dfrac{1}{2}\right)^0\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$ & $\dfrac{3!}{0! (3!)} = 1$ & $\dfrac{1}{8}$ \\
}
\caption{}
\end{table}
..and the above produces:

With \vphantom to add more space between the rows:
\begin{table}[htpb]
\centering
\pgfplotstabletypeset[
col sep=&,
row sep=\\,
string type,
columns = {x, N-x, pq, binom,prob},
columns/x/.style={column name={\boldmath{$x$}\unboldmath},
column type={@{}M{.05\dimexpr\textwidth-6\tabcolsep}}} ,
columns/N-x/.style={column name={\boldmath{$N-x$}\unboldmath},
column type={M{.1\dimexpr\textwidth-6\tabcolsep} }} ,
columns/pq/.style={column name={\boldmath{$p^{x}\ q^{N - x}$}\unboldmath},
column type={M{.3\dimexpr\textwidth-6\tabcolsep} }} ,
columns/binom/.style={column name={\boldmath{$\dfrac{N!}{x! (N - x)!}$}\unboldmath},
column type={M{.3\dimexpr\textwidth-6\tabcolsep}@{} }} ,
columns/prob/.style={column name={\boldmath{$P(x)$}\unboldmath},
column type={M{.1\dimexpr\textwidth-6\tabcolsep}@{} }} ,
every head row/.style=
{before row=\toprule\toprule, after row=\midrule},
every last row/.style=
{after row=\bottomrule\bottomrule},
]
{
x & N-x & pq & binom & prob \\
{}3 & 0 & $\Large \vphantom{\left(\dfrac{1}{2}\right)}$ $\left(\dfrac{1}{2}\right)^3\left(\dfrac{1}{2}\right)^0 = \dfrac{1}{8}$ & $\dfrac{3!}{3! (0!)} = 1$ & $\dfrac{1}{8}$ \\
\midrule
{}2 & 1 & $\Large \vphantom{\left(\dfrac{1}{2}\right)}$ $\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}\right)^1 = \dfrac{1}{8}$ & $\dfrac{3!}{2! (1!)} = 3$ & $\dfrac{3}{8}$ \\
\midrule
{}1 & 2 & $\Large \vphantom{\left(\dfrac{1}{2}\right)}$ $\left(\dfrac{1}{2}\right)^1\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{8}$ & $\dfrac{3!}{1! (2!)} = 3$ & $\dfrac{3}{8}$ \\
\midrule
{}0 & 3 & $\Large \vphantom{\left(\dfrac{1}{2}\right)}$ $\left(\dfrac{1}{2}\right)^0\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$ & $\dfrac{3!}{0! (3!)} = 1$ & $\dfrac{1}{8}$ \\
}
\caption{}
\end{table}
and the above produces:

\toprule,\midruleand\bottomruleinstead of\hlinefor the row lines, I dont have the issues with the lines being very close one to another. But still if someone has a way to change column width and row height I'll be interested. – joseldsm Oct 12 '18 at 12:09\arraystretchmacro is used with tabular and array. See what\def\arraystretch{2}does to\pgfplotstabletypeset. You can also add struts (e.g.\rule{0pt}{1cm}). – John Kormylo Oct 13 '18 at 14:07\def\arraystretch{2}does work for the row height. Thank you ! – joseldsm Oct 15 '18 at 07:38