How to open a PopupMenuButton?

I think it would be better do it in this way, rather than showing a PopupMenuButton

void _showPopupMenu() async {
  await showMenu(
    context: context,
    position: RelativeRect.fromLTRB(100, 100, 100, 100),
    items: [
      PopupMenuItem<String>(
          child: const Text('Doge'), value: 'Doge'),
      PopupMenuItem<String>(
          child: const Text('Lion'), value: 'Lion'),
    ],
    elevation: 8.0,
  );
}

There will be times when you would want to display _showPopupMenu at the location where you pressed on the button
Use GestureDetector for that

final tile = new ListTile(
  title: new Text('Doge or lion?'),
  trailing: GestureDetector(
    onTapDown: (TapDownDetails details) {
      _showPopupMenu(details.globalPosition);
    },
    child: Container(child: Text("Press Me")),
  ),
);

and then _showPopupMenu will be like

_showPopupMenu(Offset offset) async {
    double left = offset.dx;
    double top = offset.dy;
    await showMenu(
    context: context,
    position: RelativeRect.fromLTRB(left, top, 0, 0),
    items: [
      ...,
    elevation: 8.0,
  );
}