As the electronegativity of the central atom increases the bond angle also increases. So, ${\mathrm{NH}}_3$ will have high bond angle than ${\mathrm{PH}}_3$. The bond angle of ${\mathrm{NH}}_4^{+}$ is highest because of its symmetrical geometry, also bond pair and lone pair on $\mathrm{N}$ of ${\mathrm{NH}}_3$ suffer repulsion, which decreases its bond angle in comparison to ${\mathrm{NH}}_4^{+}$. Thus, the correct order of bond angle is

${\mathrm{PH}}_3<{\mathrm{NH}}_3<{\mathrm{NH}}_4^{+}$

But in the given question none of the option is correct because there is no $\mathrm{H}-\mathrm{X}-\mathrm{H}$ bond angle in ${\mathrm{BF}}_3$