Well, you are correct in the sense that the MTU represents the maximum payload of the frame. That is to say this is the maximum payload that an interface can transmit over the data link layer (Layer 2) at one time. In this regard the PDU is similar as this is the total size of the largest frame transmitted on the wire. Thus, the PDU is equal to the MTU plus the layer 2 frame header size.
Now if you are thinking of payload in terms of actual usable data this is even smaller still than the MTU as within the 1500-byte layer 2 payload of the frame the data will be further encapsulated by several higher-level protocols each of which taking up a portion of the payload available with their own headers containing essential metadata.
For instance with a 1518 byte ethernet frame on the wire carrying a TCP packet you will have the following:
- Layer 2 Ethernet Frame: Header 18 bytes - Payload 1500 bytes
- Layer 3 IP Datagram: Header 20-60 bytes - Payload 1440-1480 bytes
- Layer 4 TCP Packet: Header 20-60 bytes - Payload 1380-1460 bytes
So, the actual payload could be at most 1460 bytes and as little as 1380 bytes of the full 1518 bytes on the wire. Also, this covers the usual straight forward case where the protocols are all nested neatly inside a lower-level protocol that supports it natively. Granted the maximum size for the TCP header assumes all options are present at their maximum size this could not actually happen in a genuine packet.
Other protocols do have different overheads for example UDP has a significantly smaller fixed header of 8 bytes. Of course, the reason why TCP requires more metadata is due to all the additional features it supports so there are trade-offs for reduced overheads.
Also note this all assumes a standard Ethernet frame as Ethernet too has extensions that will increase the PDU if used as the frame header will be larger, VLANs add 4 bytes for example.