The nested loop is taking all integers from 2 to $2\ast \log_2 n$ take all their non-multiples before $n$, and make the corresponding entry in $A$ as $1$. For example, for $2$, and $n = 10, A[3], A[5], A[7]$, and $A[9] $are made $1$. Similarly for $3$, $4$, ... till $2\ast \log n$. So, if any entry $A[p]$ is $1$ means it must be a multiple of $2, 3, .... 2\ast \log_2 n$, which is $\left ( 2 \log n \right )!$ and is greater than $n$. So, for no index $p$, $A[p]$ will be $0$. So, answer is **D**.

Suppose the line

A[j] = A[j] || (j%k);

is replaced with

A[j] = A[j] || !(j%k);

Now, the nested loop is taking all integers from $2$ to $\log_2 n$, take all their multiples before $n$, and make the corresponding entry in $A$ as $1$. For example, for $2$, and $n = 10, A[4], A[6], A[8]$ and $A[10]$ are made $1$. Similarly for $3, 4, ...$ till $2\ast \log n$ . So, for all non-prime indices of $A$, we will have a $1$, and for prime indices we have a $0$. And we print $i$ if $A[j]$ is $0$ meaning $j$ is prime.