Query SQL to variable error but echo is working

I would like why I’m unable to load a row result from mysql db

This code works and result the echo: 180(value stored in PT):

$sql = "SELECT * FROM `user` WHERE pt;";
$result = mysqli_query($conn, $sql);
if (!$result) exit("The query did not succeded");
else {
    while ($row = mysqli_fetch_array($result)) {
        echo $row['pt'];
    }
}

this code result in nothing:

$sql = "SELECT * FROM user WHERE pt=?"; // SQL with parameters
$stmt = $conn->prepare($sql); 
$stmt->bind_param("i", $pt);
$stmt->execute();
$result = $stmt->get_result(); // get the mysqli result
$usert = $result->fetch_assoc(); // fetch the data   
echo $usert

testing SELECT * FROM user WHERE pt=? in phpmyadmin

1 errors were found during analysis.

Variable name was expected. (near “?” at position 38) SQL Command:
Copy Documentation

SELECT * FROM user WHERE en=? LIMIT 0.25

MySQL Messages : Documentation

#1064 – Do you have a syntax error in your SQL next to ‘? LIMIT 0.25’ on line 1

where did i go wrong and why?