Mass of the monkey, $m=40 \mathrm{~kg}$
Maximum tension that the rope can bear, $T_{\max }=600 \mathrm{~N}$
In option (a), acceleration of monkey, $a=6 \mathrm{~m} / \mathrm{s}^{2}$ (upward)
$\therefore$ By equation of motion,
$$
\Rightarrow \quad \begin{array}{ll}
T & -m g=m a \\
T & =m(g+a) \\
& =40(10+6)=640 \mathrm{~N}
\end{array}
$$
Since, $T>T_{\text {max }}$, hence in this case, rope will break. Therefore, option (a) is correct.
In option (b), acceleration of monkey, $a=4 \mathrm{~m} / \mathrm{s}^{2}$ (downward)
$\therefore$ By equation of motion,
$$
T=m(g-a)=40(10-4)=240 \mathrm{~N}
$$
Since, $T < T_{\max }$, then the rope will not break.
In option (c), monkey climbs up with uniform speed, $v=5 \mathrm{~m} / \mathrm{s}$
$\therefore$ Acceleration, $a=0 \mathrm{~m} / \mathrm{s}^{2}$
Hence, by equation of motion, $T=m g$
$$
=40 \times 10=400 \mathrm{~N}
$$
Since, $T < T_{\max }$, hence rope will not break.
In option $(\mathrm{d})$, acceleration, $a=g$
$$
\therefore \quad T=m(g-a)=m(g-g)=0 \mathrm{~N} \text {. }
$$
Since, $T < T_{\max }$, the rope will not break.
Therefore, only option (a) is correct.