How insert select multiple data into db, with same reference ID for all rows

No offense, but there is just so much wrong with your code that nothing is worth saving. Let me show you some of the main problems in your code by writing a new one as an example.

$host = 'localhost';
$db = 'test';
$user = 'username';
$pass = '1234';
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$link = new mysqli($host, $user, $pass, $db);
$link->set_charset('utf8mb4');

if (isset($_POST["submit"])) {
    $stmt = $link->prepare('INSERT INTO lesson_school_lessonstocourses SET date=current_timestamp(), progID=?, lessonID=?, comments=?');
    $stmt->bind_param('sss', $_POST['progID'], $lesson, $_POST['comments']);
    foreach ($_POST['lessonID'] as $lesson) {
        $stmt->execute();
        $last_id = $link->insert_id;
        echo "New record created successfully<br>";
    }
}

1. Always use prepared statements. Your code is vulnerable to SQL injection and you should always use parameterized prepared statements instead of manually building your queries. As an added benefit you only need to prepare the query once and you can execute it multiple times with different data which should speed up your code.

2. Enable mysqli error reporting. Use mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); and you won't have to check the return value of mysqli functions.

3. Forget about data sanitization. All it means is that you are damaging your data. Never sanitize data unless you want it broken. To prevent SQL injections use prepared statements and to prevent XSS see https://paragonie.com/blog/2015/06/preventing-xss-vulnerabilities-in-php-everything-you-need-know and How to prevent XSS with HTML/PHP?

Use a prepared statement and do the INSERT queries inside the loop.

$stmt = $link->prepare("INSERT INTO lesson_school_lessonstocourses SET
    date = current_timestamp(),
    progID = ?,
    lessonID = ?,
    comments = ?");
$value = null;
$stmt->bind_param("iis", $progID, $value, $comments);
foreach ($lessonIDs as $value) {
    if ($stmt->execute()) {
        echo "New record inserted successfully";
        $last_id = $stmt->insert_id;
    } else {
        echo "<p>Error adding lesson: " . $stmt->error . "</p>";
    }
}

Note that $last_id will just be the ID of the last lesson that was inserted. If you need all of them you should put them in an array instead of a single variable.