Centering a column in align or alignat with desired spacing

Question

Initially, I had the code

\begin{align*}
|a_d|_p & = 1 \\
|a_i|_p & \leq 1
\end{align*}

which produced the result below:

This is undesirable because the vertical bars | on the left don't line up. Then I discovered the alignat environment, and tried the code

\begin{alignat}{3}
| & a_d & |_p & & & = 1 \\
| & a_i & |_p & & & \leq 1
\end{alignat}

which produced the result below:

This is better, but it's still not great because now the a_d and a_i don't line up. Ideally, I'd want just the column with the a_d and a_i to be centered. Then, I looked at the answer to this question, and tried the code

\begin{alignat*}{3}
| & \omit\hfill $a_d$ \hfill & |_p & & & = 1 \\
| & \omit\hfill $a_i$ \hfill & |_p & & & \leq 1
\end{alignat*}

which produced the result below:

This is better in some respects because each of the vertical bars line up, as do the a_d and a_i, but now the right-hand vertical bars |_p are spaced too far to the right. Any suggestions?

---

UPDATE: After reading the code in Bernard's suggestions, along with some information about \makemathbox (from mathtools), \newlength, and \settowidth, I came up with the following code, which seems to do what I want. However, am I doing anything that is deprecated?

\newlength{\myl}
\settowidth{\myl}{$a_d$}

\begin{align*}
|a_d|_p & = 1 \\
| \mathmakebox[\myl]{a_i} |_p & \leq 1
\end{align*}

Answer 1

Here are two solutions, using mathtools, and makebox for one of them. For the second one , with eqparbox , more automated. I define a \varpabs command which uses a tag: all \varpabs which share the same tag will have the same width – that of the widest argument:

\documentclass{article}

\usepackage{mathtools}
\usepackage{etoolbox}
\usepackage{makebox, eqparbox}
\DeclarePairedDelimiterXPP{\pabs}[1]{}\lvert \rvert{_p}{\ifblank{#1}{\:\cdot\:}{#1}}
\DeclarePairedDelimiterXPP{\varpabs}[2]{}\lvert\rvert{_p}{\eqmakebox[#1]{$\displaystyle#2$}}

\begin{document}

\begin{align*}
  \pabs{a_d} & = 1 \\
  \pabs{\makebox*{$ a_d $}{$ a_i $}} & \leq 1
\end{align*}

\begin{align*}
  \varpabs{A}{a_d + c} & = 1 \\
  \varpabs*{A}{ \frac{a_i }{b_i}} & \leq 1
\end{align*}

\end{document}