In relational logic, why is $p(x) \Longrightarrow \forall(x).p(x)$ contingent, but $p(x) \Longrightarrow \exists(x).p(x)$ valid?
How would you interpret $p(x)$? Is $p(x)$ a ground atom, where $x$ is some object? If so, then $p(x) \Longleftrightarrow \exists(x).p(x)$, which explains the results.
However, to my knowledge, $x$ is reserved for variables. If $x$ is a free variable, we should wrap the relation $p(x)$ with a universal quantifier $\forall$. Is this true?
I assume you mean:
Why is $p(x) \Longrightarrow \forall(x) p(x)$ contingent, but $p(x) \Longrightarrow \exists(x) p(x)$ valid?
First, let's get clear on the notions of 'contingengy' and 'validity' here, especially since both sentences contain a free variable.
So, let's say that a formula $\phi$ is contingent if there is some interpretation $I$ and variable assignment $v$ such that when we assign objects (from the domain of $I$) to the free variables in $\phi$, the resulting sentence is true under $I$, and there is also some interpretation $I'$ and variable assignment $v'$ such that when we assign objects (from the domain of $I$) to the free variables in $\phi$, the resulting sentence is false under $I$. A formula is valid if for any interpretation $I$ and any variable assignment $v$ it is true that when we assign objects (from the domain of $I$) to the free variables in $\phi$ in accordance with $v$, the resulting sentence is true under $I$.
Now:
$p(x) \Longrightarrow \forall(x) p(x)$ is contingent because
If you take a domain with exactly one object, and say it has property $p$, then the $x$ in the antecedent $p(x)$ will have to be that one object, and thus be true, while the consequent $\forall(x) p(x)$ is also true, meaning that the whole conditional becomes true.
And if you take a domain with two objects (say $a$ and $b$), and one object $a$ has property $p$, and the other $b$ does not, and you take a variable assignment that assigns $a$ to $x$, then $p(x)$ would be true, but $\forall(x) p(x)$ would be false, and hence the conditional is false.
On the other hand, $p(x) \Longrightarrow \exists(x) p(x)$ is valid, because for any interpretation $I$ and variable assignment $v$: if $p(x)$ is true under $I$ and $v$, then there must be some object int he domain with property $p$, and hence $\exists(x) p(x)$ will also be true under $I$ and $v$. Hence the conditional is always true under any $I$ and $v$, meaning that the formula is valid.
Now, to answer your other question: is the validity of $p(x) \Longrightarrow \exists(x) p(x)$ because the $p(x)$ without the quantifier implicitly assume an existential? And as such, would it be the case that $p(x) \Longleftrightarrow \exists(x) p(x)$ is valid?
No. Consider if we go the other way:
$ \exists(x) p(x) \Longleftrightarrow p(x)$
Is that valid? No, because I can pick the earlier domain with two objects (say $a$ and $b$), where one object $a$ has property $p$, and the other $b$ does not, and take a variable assignment that assigns $b$ to $x$. Then $\exists(x) p(x)$ would be true, but $p(x)$ would be false, and hence the conditional would be false under that $I$ and $v$. So, this is not a valid formula, and consequently $p(x) \Longleftrightarrow \exists(x) p(x)$ is not a valid formula either.
In other words, having the $p(x)$ without any quantifiers does not work like an existential, nor like a universal.