How to make this search query work together with these placeholders and what is wrong

The first query work when I remove the placeholder :p and replace with ? and at the $stmt->execute(array(':s' => $s)); replace with $stmt->execute([$s]); while the code after if(!empty($search) { is not working, look below

    $mydb = new Mydb;
    //a possessed id
    $s = 1;
    $q = 'SELECT name, street FROM mydata WHERE possessed_id = :p ';
    $search = $_POST["search"]["value"];
    if(!empty($search) {
        $q .= 'AND (name LIKE :n OR street LIKE :s)';
    }
    $stmt = $mydb -> prepare($q);
    $stmt->bindValue(':n', '%'.$search.'%', PDO::PARAM_STR);
    $stmt->bindValue(':s', '%'.$search.'%', PDO::PARAM_STR);
    
    $stmt->execute(array(':s' => $s));
  

the goal is to display all data in a table and allow search

Answer

When $search is empty, you have the parameter :p in your query but you try to bind the :n & :s parameters. When $search is not empty, you have the 3 parameters (:p, :n & :s) but you still don’t bind :p.
In any case you have a different number of declared and bound parameters.

You should take care of :p and only bind the other parameters when they exist in the query:

$q = 'SELECT name, street FROM mydata WHERE possessed_id = :p ';
$search = $_POST["search"]["value"];

// Default parameters
$parameters = [
    'p' => $s
];

// Add ":n" & ":s"
if('' !== $search) {
    $q .= 'AND (name LIKE :n OR street LIKE :s)';
    $parameters['n'] = '%' . $search . '%';
    $parameters['s'] = '%' . $search . '%';
}

$stmt = $mydb->prepare($q);
// Send the parameters
$stmt->execute($parameters);

I assumed a corrected version of your parameters and their value and used only 1 way to bind the parameters to the query with an array during the execute() call.

I also removed the empty() call (here‘s why).