Step 1: Bring in the drift variance formula.
The variance added to an allele frequency each generation by drift is
\[ \text{Var}(\Delta p) = \frac{p(1-p)}{2N} \]
where $N$ is the population size. This tells us directly that a smaller $N$ gives a bigger variance, meaning bigger random jumps in $p$ from one generation to the next.
Step 2: Plug in the two population sizes.
For M, $N=100$, so the variance term has $2N=200$ in the denominator. For N, $N=1000$, so the denominator is $2N=2000$, ten times larger. This means the per generation variance in $p$ for M is about ten times that for N.
Step 3: Add up drift over 100 generations.
Repeated random steps accumulate, so after $100$ generations the spread of possible $p$ values around $0.5$ is much wider for M than for N. In population M, $p$ is far more likely to have wandered close to $0$ or $1$ by now.
Step 4: Translate spread in $p$ into heterozygosity.
$2pq$ peaks at $p=0.5$ and falls as $p$ moves toward either extreme. Since M's $p$ has likely wandered further from $0.5$ than N's, M's expected $2pq$ is lower.
Step 5: Address the directional options.
Options (A) and (B) claim $p$ ends up on a particular side of $q$ in one population. But the drift process is symmetric around $0.5$, so neither direction is favored in either population, which rules both out.
Step 6: Conclude.
The smaller population loses heterozygosity faster.
\[ \boxed{2pq_M < 2pq_N} \]