why is the Program Address Register in 8051 connected to a 8-bit bus?

In this Diagram found in Wikipedia, both ROM and Program Address Register are connected to a 8-bit bus, but I thought in 8051, Internal Rom uses 16-bit adrress. So why is the Program Address Register connected to a 8-bit bus?

Since we are not the ones who designed this architecture, we can only guess or think of reasons. If you actually want to know why, you need to find the ones who made the decision.

The 8051 multiplexes the lower address byte and the data on P0, and the upper byte of the address can go through P2. Because both bytes can be transferred at different times, an 8-bit bus suffices.

This design decision could be made to reduce silicon needs, in terms of transistors and wires. Silicon area was expensive that time.

There might be a history in the issue, too. The 8051 series followed the MCS48 (8021, 8022, 8048, etc) that had fewer address bits.

Think about the time this controller was designed, and the target market. External RAM or memory-mapped peripherals might only need 8 bits of address, just because it was enough, or too costly to use more.

Another reason might be over-simplification in the diagrams, that additionally may base one on the other. The program address register might use two 8-bit busses, each connected to one port driver block. And some "wise" guy simply put them together.