Hi, i am try to display and insert info on a person into my database but i keep getting this error whenever i enter the info

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'join (FName, LName) VALUES ('flsdjflsjfl','jsdlfjslfjvlj')' at line 1

include ("include.php");
	$display_block = "
	<form method=\"post\" action=\"".$_SERVER["PHP_SELF"]."\">
	<p><strong>First/Last Names:</strong><br/>
	<input type=\"text\" name=\"FName\" size=\"30\" maxlength=\"75\">
	<input type=\"text\" name=\"LName\" size=\"35\" maxlength=\"50\"></p>
	<input type=\"text\" name=\"Street\" size=\"50\"></p>
	<input type=\"text\" name=\"City\" size=\"30\" maxlength=\"50\">
	<input type=\"text\" name=\"State\" size=\"5\" maxlength=\"2\">
	<input type=\"text\" name=\"Zip\" size=\"10\" maxlength=\"10\"></p>
	<p><strong>Street Type:</strong><br/>
	<input type=\"radio\" name=\"st_type\" value=\"home\" checked> home
	<input type=\"radio\" name=\"st_type\" value=\"work\">work</p>
	<input type=\"text\" name=\"Phone\" size=\"30\" maxlength=\"25\">
	<input type=\"radio\" name=\"tel_type\" value=\"home\" checked> home
	<input type=\"radio\" name=\"tel_type\" value=\"work\" checked>work</p>
	<p><strong>Email Address:</strong><br/>
	<input type=\"text\" name=\"Email\" size=\"30\" maxlength=\"150\">
	<input type=\"radio\" name=\"email_type\" value=\"home\" checked>home
	<input type=\"radio\" name=\"email_type\" value=\"work\" checked>work</p>
	<p><input type=\"submit\" name=\"submit\" value=\"Add Advocate\"></p>
} else if($_POST){
	if(($_POST["FName"] == "") || ($_POST["LName"] == "")){
		header("location: addadvocate.php");
	//connect to database
	$add_master_sql = "INSERT INTO join (FName, LName) 
						VALUES ('".$_POST["FName"]."','".$_POST["LName"]."')";
	$add_master_res = mysqli_query($mysqli, $add_master_sql)
					  or die(mysqli_error($mysqli));
	if(($_POST["Street"]) || ($_POST["City"]) || ($_POST["State"])
		|| ($_POST["Zip"])){
		$add_street_sql = "INSERT INTO join (Street, City, State, Zip)
							VALUES ('".$_POST["Street"]."', '".$_POST["City"]."', 
							'".$_POST["State"]."', '".$_POST["Zip"]."')";
		$add_street_res = mysqli_query($mysqli, $add_street_sql)
						   or die(mysqli_error($mysqli));
		$add_tel_sql = "INSERT INTO join (Phone) VALUES
		$add_tel_res = mysqli_query($mysqli,$add_tel_sql)
					   or die(mysqli_error($mysqli));
		$add_email_sql = "INSERT INTO join (Email) VALUES ('".$_POST["Email"]."')";
		$add_email_res = mysqli_query($mysqli, $add_email_sql)
						 or die(mysqli_error($mysqli));
	$display_block = "<p>Your entry has been added. Thank you! Would you 
	like to <a href=\"addadvocate.php\">add another</a>?</p>" ;
<title>Add an Entry</title>
<h1>Add an Entry</h1>
<?php echo $display_block; ?>

I think, join is mysql reseverd keyword so rename it and then try again.
Also you must write only one query. Here you are writing so many insert queries for one record (I GUESS). Build single insert query.

Yes urtrivedi is right.
you can always use `(escape) character around table name to avoid it.