The user process generated a virtual address in binary form is
0001 0001 0001 0010 0011 0100 0101 0110
1 1 1 2 3 4 5 6
$$\begin{array}{|c|c|c|} \hline 0001 & 0001 & 0001 & 0010 & 0011 & 0100 & 0101 & 0110 \\ \hline 1 & 1 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \end{array}$$
Now it is given the page size is $2^{12}$ ,
we take lower order 12 bits from above binary form of virtual address, that is 0100 0101 0110 are used as the displacement into the page .
The page table size is $2^{20}$, so remaining 20 bits ( 0001 0001 0001 0010 0011 ) are used as the displacement in the page table.