The person went out between $4$ P.M. and $5$ P.M., hence hour hand would be between $4$ and $5$ on the clock. On her return (between $5$ P.M. to $6$ P.M.) she observed that position of the hour hand has been replaced by minute hand and position of the minute hand has been replaced by hour hand, so minute hand would have been between $5$ and $6$ (when he went to friends' house) on the clock.
$\underline{\textbf{Observations}}$
At $4$ P.M., hour hand would be pointing to $4$ and minute hand pointing to $12$ on the clock.
If minute hand is pointing to $5$ on the clock then it would be making $150^{\circ}$ from $12$ on the clock ($\because$ angle between every two consecutive numbers on the clock is $\frac{360^{\circ}}{12} = 30^{\circ}$)
When minute hand would be between $5$ and $6$ on the clock, making (say) $150^{\circ}+\theta_1^{\circ}$ from $12$ on the clock, then hour hand would have moved by (say) $\theta_2^{\circ}$ from $4$ on the clock, where $\theta_1^{\circ}, \theta_2^{\circ} \le 30^{\circ}$.
When minute hand covers $360^{\circ}$ then hour hand covers $30^{\circ}$. This implies minute hand covers $12\times$ of what hour hand covers.
Suppose hour hand covers $x^{\circ}$, and in that much time minute hand covers $y^{\circ}$ ($x^{\circ} \rightarrow y^{\circ}$) then we can say that $12x^{\circ} = y^{\circ}$. In next section, we will use this particular observation directly.
$\underline{\textbf{Calculations}}$
When hour hand covered $\theta_{2}^{\circ}$, minute hand covered $150+\theta_1^{\circ}$.
$\theta_2^{\circ} \rightarrow 150^{\circ} + \theta_1^{\circ}$
$$\begin{align} \implies 12\theta_2^{\circ} &= 150^{\circ} + \theta_1^{\circ} \tag{1}\end{align}$$
When hour hand replaces the position of minute hand, then hour hand need to cover $(30^{\circ}-\theta_2^{\circ}) + \theta_1^{\circ}$. When minute hand replaces the position of hour hand, then minute had need to cover $360^{\circ}-((30^{\circ}-\theta_2^{\circ}) + \theta_1^{\circ})$.
$(30^{\circ}-\theta_2^{\circ}) + \theta_1^{\circ} \rightarrow 360^{\circ}-((30^{\circ}-\theta_2^{\circ}) + \theta_1^{\circ})$
$$\begin{align} \implies 12(30^{\circ}-\theta_2^{\circ}) + 12\theta_1^{\circ} &= 360^{\circ}-((30^{\circ}-\theta_2^{\circ}) + \theta_1^{\circ}) \tag{2}\end{align}$$
Solving $(1)$ and $(2)$, we get $\theta_1^{\circ} = 11.118^{\circ}; \ \theta_2^{\circ} = 13.426^{\circ}$.
We have been asked to find the time when she went out to chat. For that we still need to know exactly how many minutes past $4$ she went out. As we can observe in the diagram above that it would be $\text{MinutesPastIn}(\theta_1^{\circ}) +25$ minutes.
For minute hand, in $1^{\circ}$, $\frac{60}{360^{\circ}} = \frac{1}{6}$ min passes by. So for $\theta_1^{\circ}=11.1185^{\circ}$, $\frac{11.1185^{\circ}}{6} = 1.85$ minutes passes by. This is same as $26$ and $0.85$ minutes past $4$ P.M., or $\fbox{$26$ and $\frac{122}{143}$ minutes past $4$ P.M.}$.
$\textbf{Option (B) is correct}$