从我看到的示例输入和期望的输出来看，不需要preg_replace_callback()的开销，只要星号后面跟有4个或更多的数字或连字符，可变长度的前瞻就允许您一次用星号替换一个字符。
代码：（Demo）

$inputs = [
    "John's Mobile number is 8767484343 and he is from usa.",
    "John's Mobile number is +918767484343 and he is from india.",
    "Sunny's Mobile number is 08767484343 and he is from Pimpri-Chinchwad, india.",
    "Rahul's Mobile number is 1800-190-2312 and he is from india."
];

var_export(preg_replace('~[+\d-](?=[\d-]{4})~', '*', $inputs));

输出：

array (
  0 => 'John\'s Mobile number is ******4343 and he is from usa.',
  1 => 'John\'s Mobile number is *********4343 and he is from india.',
  2 => 'Sunny\'s Mobile number is *******4343 and he is from Pimpri-Chinchwad, india.',
  3 => 'Rahul\'s Mobile number is *********2312 and he is from india.',
)

我可以想象一些我的代码片段无法处理的边缘情况，但无论何时，当您处理不遵守严格格式的电话号码时，您都将陷入一个充满挑战的兔子洞。

你可以直接从你的$dynamic_var实现，例如：

$dynamic_var = "John's Mobile number is 8767484343 and he is from usa.";
$result = preg_replace_callback('/(?<=\s)(\d|-|\+)+(?=\d{4}\s)/U', function($matches) {
    return str_repeat("*", strlen($matches[0]));
}, $dynamic_var);

古老但有用...

<?php
    echo str_repeat('*', strlen("123456789") - 4) . substr("123456789", -4);
?>