Answer: A,C
Option A) At $P$, datagram size is : $1400+20(IP header)$ = $1420B$, with identification number $0x1234$
When it passes through router $R$, since the MTU of $R-Q$ is $820B$, the datagram needs fragmentation at Router $R$.
First fragment : $800B$ (data) + $20B$ (IP Header) with identification number $0x1234$.
Second fragment : $600B$ (data) + $20B$ (IP Header) with identification number $0x1234$. Hence Option A is True.
Option B) When a fragment is lost, router doesn't resend the same fragment again, that too with same Identification Number. The whole TCP segment needs to be resent again and again fragmentation happens at router $R$. The Identification number may or may not be same as $0x1234$.
When sending an identical copy of an earlier datagram, a
host MAY optionally retain the same Identification field in
the copy.
Source: RFC 1122
This makes Option B False and Option C True.
Option D) Destination Port is a $16 bits$ number, just after the first $16 bits$ of Source Port number. It is included in TCP header of every segment. During reassembly, which happens at the destination, the TCP destination port be determined by analyzing any of the fragments and not only the second fragment. Hence Option D is False.