Error on Undefined variable

I have add this direct under
$tip = ”;

if ( ! function_exists( ‘get_serach_fields_multi_select’ ) ) {

function get_serach_fields_multi_select($value) {
$selections = (array) get_option( $value[‘id’] );

we are not using $tip variable. you can remove it.

If you want to define $tip as null ($tip = '';) then you must have to define inside the function so, you should be use like below.

if ( ! function_exists( 'get_serach_fields_multi_select' ) ) {
function get_serach_fields_multi_select($value) {
$tip = '';
$selections = (array) get_option( $value['id'] );