Solution:

We are given:

$$ \text{Logical address space} = 32 \text{ KB}, \quad \text{Page Size} = 4 \text{ KB}, \quad \text{Physical Memory size} = 64 \text{ KB} $$

Step 1: Number of pages in logical address space

The number of pages is given by:

$$ \text{Number of pages} = \frac{\text{Logical Address Space}}{\text{Page Size}} $$

Substituting values:

$$ \frac{32 \ \text{KB}}{4 \ \text{KB}} = 8 $$

So, there are 8 pages.

Step 2: Number of page frames in physical memory

The number of page frames is given by:

$$ \text{Number of frames} = \frac{\text{Physical Memory Size}}{\text{Page Size}} $$

Substituting values:

$$ \frac{64 \ \text{KB}}{4 \ \text{KB}} = 16 $$

So, there are 16 page frames.

Final Answer:

The number of pages in the logical address space and the number of page frames in the physical memory are:

$$ (8, \ 16) $$

Hence, the correct option is Option C.