PHP Parse error: syntax error, unexpected $end under ?>

I am trying to get this php code to run and I am getting an error that says "PHP Parse error: syntax error, unexpected $end in /var/www/TestSite1/betting.php on line 54", I checked multiple times that I closed all of the brackets, quotations, and semicolons. Any help would be greatly appreciated.

<form action="index.php" method="post">
       Person: <select name="person">
    <option value="Karl">Karl</option>
    <option value="Erik">Erik</option>
    <option value="Patrick">Patrick</option>
    </select>
       Money: <input type="text" name="am"><br>
       Pass: <input type="text" name="pass"><br>
    <input type="submit">
    </form>
    <?php
    $person=$_POST["pass"];
    $money=$_POST["am"];

       class MyDB extends SQLite3
    {
      function __construct()
      {
        $this->open('betting.db');
      }
    }
    $db = new MyDB();
    if(!$db){
      echo $db->lastErrorMsg();
    } else {
      echo "Opened database successfully<br>";
    }
    if ($passwd == '3nt3rpr1s3'){
      print "Correct Password<br>";
       $sql =<<<EOF
         UPDATE people set money = "$money" where person=$pname;
         EOF;
    } else {
      print "\nIncorrect Password!";
    }
    $ret = $db->exec($sql);
    if(!$ret){
      echo $db->lastErrorMsg();
    } else {

    }

       $sql =<<<EOF
      SELECT * from people;
    EOF;
    $ret = $db->query($sql);
    while($row = $ret->fetchArray(SQLITE3_ASSOC) ){
      echo "<br>ID = ". $row['id'] . "\n";
      echo "<br>Person = ". $row['person'] . "\n";
      echo "<br>Money = ". $row['money'] . "\n";
    }
    $db->close();
    ?>