The statement $P \to \left( {q \to p} \right)$ is equivalent to
A.$P \to \left( {p \wedge q} \right)$
B.$P \to \left( {p \leftrightarrow q} \right)$
C.$P \to \left( {p \to q} \right)$
D.$P \to \left( {p \vee q} \right)$

Hint: In order to solve this particular question related to the topic of mathematical reasoning, we need to follow all the rules of deductive reasoning and hence we will get our final result for this question.

Complete step-by-step answer:
In order to solve this question, we have applied the concept of mathematical reasoning.
Mathematical Reasoning:-
A sentence is called a mathematically acceptable statement if it is either true or false but not both. It is generally denoted as an alphabet followed by a colon and then the statement. A sentence is neither imperative nor interrogative or exclamatory.
A declarative sentence containing variables is an open statement if it becomes a statement when the variables are replaced by some definite values.
There are three types of implications:
a.) “If ….... then”
b.) “Only if”
c.) “If and only if”
“If …. then” type of compound statement is called a conditional statement. The statement ‘if p then q’ is denoted by p → q or by p ⇒ q. p → q also means:
a.) p is sufficient for q
b.) q is necessary for p
c ) p only if q
d.) p leads to q
e.) q if p
f.) q when p
g.) if p then q
We know that, p → q is false only when p is true and q is false.
$P \to \left( {q \to p} \right) = \sim p \vee \left( {q \to p} \right) = \sim p \vee \left( { \sim q \vee p} \right)$
$ = \sim p \vee p \vee q = T \vee q = True$
Since $p \vee \sim p$ is always True.
$p \to \left( {p \vee q} \right) = \sim p \vee p \vee q$
$ = T \vee q$
= True.
So, $P \to \left( {q \to p} \right) = P \to \left( {p \vee q} \right)$
$\therefore $ Option (d) is the correct answer for this particular question.


Note: If a statement is denoted by ‘p’ then its negation is denoted by '∼ p'.
It is very important to note that p → q is false only when p is true and q is false for solving this question.