 Solution

For a fixed prime $p\in [P,2P]$, the grid $a/p$ has circular spacing $1/p\ge 1/(2P)$. Consequently an arc of length $\Delta =1/P$ contains at most three points of this grid, including endpoints. The standard Chebyshev estimate $\pi (t)\ll t/\log t$ for the prime-counting function givesSince $N\le P$, the local-multiplicity large sieve yields the prime-denominator large sieve:By contrast, distinct reduced fractions with denominators at most $2P$ have circular distance at least $1/(4P^2)$: their difference, even after subtraction of an integer, has a nonzero integer numerator over denominator $pp\prime$. Applying part (a) alone gives only $(4P^2+2\pi N)E\ll P^{2}E$. The local-multiplicity argument saves a factor of $\log P$.