How can I handle both 'pages[0]' and 'parentPage'?

That's a great solution—much more concise than mine.
I'll see if I can incorporate it into my code.
Thank you very much.

Use something like this

This keeps everything in one code path and avoids splitting your logic.

var d = app.activeDocument;
var p = item.parentPage;

// fallback if item is not on a page
if (!p) {
    try {
        p = item.parent.parentPage;
    } catch(e) {}
}

// final fallback (e.g. pasteboard items)
if (!p) {
    p = d.pages[0]; // or skip entirely depending on your logic
}

// now safe to use
var pb = p.bounds;

If you want to be more precise (and avoid defaulting to page 0), you can resolve the page from the object’s position instead
 

function getPageFromItem(item) {
    var d = app.activeDocument;

    if (item.parentPage) return item.parentPage;

    var gb = item.geometricBounds;
    var y = gb[0];
    var x = gb[1];

    for (var i = 0; i < d.pages.length; i++) {
        var p = d.pages[i];
        var b = p.bounds;

        if (y >= b[0] && y <= b[2] && x >= b[1] && x <= b[3]) {
            return p;
        }
    }

    return null;
}

Then

var p = getPageFromItem(item);
if (!p) return; // or handle as needed

var pb = p.bounds;

And yes, p[0] / p[1] refers to page order in the spread (typically left/right), but that’s not reliable logic for identifying the “current” page of an item. I try to never infer page ownership from index and try to always resolve it from the item itself.

What your issue so far is, I think::

parentPage is correct but not guaranteed to exist
pages[0] is too generic a fallback, probably not a solution
From most I’ve seen they try to get the page dynamically from the item

That way it keeps a single, consistent code path without branching the entire script.