(E) For reaction $(d)$: $PCl_{3(l)} + 3H_2O_{(l)} \to 3HCl_{(aq)} + H_3PO_{3(aq)}$
Calculating oxidation states: $P$ in $PCl_3$ is $+3$,$P$ in $H_3PO_3$ is $+3$. $Cl$ is $-1$ in both $PCl_3$ and $HCl$. $H$ is $+1$ and $O$ is $-2$ in both reactants and products.
Since there is no change in oxidation states,this is not a redox reaction.
For reaction $(e)$: $4NH_{3(aq)} + 3O_{2(g)} \to 2N_{2(g)} + 6H_2O_{(g)}$
Calculating oxidation states: $N$ in $NH_3$ is $-3$,$N$ in $N_2$ is $0$. $O$ in $O_2$ is $0$,$O$ in $H_2O$ is $-2$.
Here,$N$ is oxidized ($-3$ to $0$) and $O$ is reduced ($0$ to $-2$).
Therefore,this is a redox reaction.
$O_2$ acts as the oxidising agent and $NH_3$ acts as the reducing agent.