Q:

Make jquery plugin by function in result

I am trying to create a jQuery plugin.
$.fn.examplePlugin = function(callback) {
    $.item = $(this);
    $(this).hide('fast',function () {
        callback('bla');
    });

    function show(a) {
       alert(a);
    }    

}

and use
$('form').examplePlugin(function(data)) {
    this.show(data); // need to return alert 'bla'
});

need to return alert 'bla'

A:

After your edits, your code almost does what you mean to do. There are only a couple of problems:

What is the purpose of this line?
$.item = $(this);

You're creating a field in the jQuery object (alias $) that is referencing an element. If you call your plugin N times, that field will be overwritten N times. Whatever you were trying to accomplish, this is not the way to do it.
Your function show is only available inside the closure function(callback) { ... }, you can't access it from outside. Calling this.show will do nothing, and you shouldn't try to add members to this because it refers to a DOM element.
For a "proper" way of doing it, I'd suggest reading some tutorials on plugin authoring, but as a quick workaround, why don't you pass the show function as argument to the callback also?
$(this).hide('fast',function () {
    callback(show, 'bla');
});

...

$('form').examplePlugin(function(show, data) {
    show(data); // will perform (not return) "alert('bla')"
});

