Given:

\[ \begin{aligned} &\text{CPU registers} = 6 \\ &\text{Instruction length} = 32\ \text{bits} \\ &\text{Main memory size} = 512\ \text{KB} = 512\times 2^{10}=2^{9}\times 2^{10}=2^{19}\ \text{bytes} \end{aligned} \]

Assumption: memory is byte-addressable (each distinct address selects one byte).

Number of bits required for a memory address = \(\log_2(2^{19}) = 19\ \text{bits}\).

Number of bits to encode one register = \(\left\lceil \log_2 6 \right\rceil = 3\ \text{bits}\), so two registers need \(3+3=6\ \text{bits}\).

Total bits used by operands = \(19\ (\text{address}) + 6\ (\text{two registers}) = 25\ \text{bits}\).

Remaining bits for opcode = \(32 - 25 = 7\ \text{bits}\).

\[ \therefore\ \text{Maximum distinct instructions} = 2^{7} = \boxed{128}. \]

(Note: if addresses were word-addressable or some other addressing unit were used, the address-bit count — and hence the result — would change.)