A model in which the formula holds I'm supposed to find a model in which the formula
$$\exists x(\forall y(P(x) \land Q(x,y)))$$
holds.
Does the model below work?:
$M$:

*

*$U = \{a,b\}$,

*$P=\{a,b\}$,

*$Q=\{(a,b), (b,a)\}$.

 A: No, the interpretation you proposed does not make the formula
$$\tag{$*$}
\exists x(\forall y(P(x) \land Q(x,y)))
$$
true. Indeed, according your interpretation, there are two possibilities for $x$:

*

*either $x = a$, and then for $y = a$ we have (according your interpretation) that $P(a)$ holds, but $Q(a,a)$ does not hold;

*or $x = b$, and then for $y = b$ we have (according your interpretation) that $P(b)$ holds, but $Q(b,b)$ does not hold.

To solve the problem, and have an interpretation $I$ that makes the formula $(*)$ true, you can keep a domain $|I|$ with exactly two elements $a$ and $b$, and slightly change the interpretations of $P$ and $Q$ in $I$ as follows:

!\begin{align}
P^I &= \{a\}
\\
Q^I &= \{(a,a), (a,b)\}
\end{align}

If you look for a more "minimalist" solution, you can take an interpretation $J$ whose domain is $|J| = \{a\}$ (only one element) and the interpretations of $P$  and $Q$ are as follows:

!\begin{align}
P^I &= \{a\}
\\
Q^I &= \{(a,a)\}
\end{align}

A: You can use the Tree Proof Generator. It says that the expression
$$\tag{1}
\exists x(\forall y(P(x) \land Q(x,y)))$$
is invalid and displays:
Countermodel:
Domain: { 0 }
P:  { }
Q:  { }

If you input
$$\tag{2}
\exists x(\forall y(P(x) \land Q(x,y)))$$
instead, it also says that the expression is invalid and displays
Countermodel:
Domain: { 0 }
x:  0
P:  { }
Q:  { }

So for the $x$ that should exist you can choose $x=0$. For $y=0$  we have
$$P(0) \land Q(0,0)$$
is false.
