$4$ should be the correct answer.$$\begin{array}{|l|l|l|l|} \hline \textbf{P} & \textbf{Q} & \textbf{P} \implies \textbf{Q} & \textbf{P} \wedge ( \textbf{P} \implies \textbf{Q}) \\\hline \text{F} & \text{F} & \text{T} & \text{F} \\\hline \text{F} & \text{T} & \text{T} & \text{F} \\\hline \text{T} & \text{F} & \text{F} & \text{F} \\\hline \text{T} & \text{T} & \text{T} & \text{T} \\\hline \end{array}$$
Suppose $\left(P\wedge(P\implies Q)\right)\iff A$ (For notational convenience)
Thus for options, $(i),(ii),(iii),(iv),(v)$
If $(A\implies \text{option x})$ is a tautology.
then $P\wedge(P\implies Q)$ logically implies $\text{option x}$
else $P\wedge(P\implies Q)$ does not logically implies $\text{option x.}$
$$\small{\begin{array}{|c|c|c|cc|cc|cc|cc|cc|} \hline \textbf {P}& \ \textbf {Q} &\ \textbf{A}&\rlap{\textbf{Option(i)} }&& \rlap{\textbf{Option(ii)}}& &\rlap{\bf{Option(iii)} }&& \rlap{\bf {Option(iv)}}&& \rlap{\bf{Option(v)}} \\\hline &&& \text{False}&A \Rightarrow F & \text{Q}&A \Rightarrow Q&\text{True}&A\Rightarrow true&(P \vee Q)&A \Rightarrow (P \vee Q)&\neg Q \vee P&A \Rightarrow (\neg Q \vee P))\\\hline F&F&F& & T && T && T && T && T\\\hline F & T & F && T & &T &&T &&T&&T \\\hline T & F & F && T & &T &&T &&T&&T \\\hline T & T & T &F& F & T&T &T&T &T&T&T&T \\\hline \end{array}}$$
Answer $=4$
P.S: Blank entries in the above truth table are like don't care conditions because in those rows the value of $A$ is set to False. Hence, $(A\implies \text{Anything})$ would be set to True.