See to the following changes:

<input type='file' name='inputfile' id='inputfile'>

Here's how you should have sent the ajax request:

$(document).ready(function() {
    $('#inputfile').change(function(){
        var file_data = $('#inputfile').prop('files')[0];   
        var form_data = new FormData();                  
        form_data.append('file', file_data);
        $.ajax({
            url: "pro-img-disk.php",
            type: "POST",
            data: form_data,
            contentType: false,
            cache: false,
            processData:false,
            success: function(data){
                console.log(data);
            }
        });
    });
});

And lastly, here's how you should have processed the form data:

$src = $_FILES['file']['tmp_name'];
$targ = "../images/".$_FILES['file']['name'];
move_uploaded_file($src, $targ);

Try this:

var file_data = $('#inputfile').prop('files')[0];
var form_data = new FormData();                     // Create a form
form_data.append('inputfile', file_data);           // append file to form

$.ajax({
        url: "pro-img-disk.php",
        type        : 'post',
        cache       : false,
        contentType : false,
        processData : false,
        data        : form_data,                         
        success     : function(response){
            alert(response);
        }
 });

in php you can get the file data like:

$_FILES['inputfile']