How can we express "$P=PSPACE$" as a first-order formula?
Firstly, I want to address the comments to the question, where it was suggested that "false" expresses $P = PSPACE$ because the statement is false. While this might be a good joke, it is actually very much harmful to think this way. When we ask how to express a certain sentence in a certain formal system, we are not talking about truth values. If we were, then when someone asked "How do I write down the fact that there are infinitely many primes?" we could answer "3 + 3 = 6", but this clearly will not do. For the same reason "false" is not a valid answer to "how do I write down $P = PSPACE$?". I think Frege and Russell tried hard to teach us that lesson. Ok, now to the answer.

Let me show how to express $PSPACE \subseteq P$, the other direction is similar, and then you can put them together in a conjunction to get $PSPACE = P$. In any case, for your purposes it may be sufficient to express just $PSPACE \subseteq P$, depending on what you are doing.

Using techniques similar to those in the construction of Kleene's predicate $T$, 
we can construct a bounded quantifer formula $accept_{space}(k, m, n)$ (which thus resides in $\Sigma^0_0 = \Pi^0_0$) saying "when we run the machine encoded by $k$ and bound its space usage to $|n|^m$, the machine accepts the input $n$." Here $|n|$ is the length of $n$. An informal way of seeing that such formulas exists is this: given $k$, $m$, and $n$ we can compute primitive recursive bound on how much time and how much space we are ever going to need (i.e., at most $|n|^m$ space and at most $2^{|n|^m}$ time). We then simply search through all possible execution traces which are within the computed bounds--such a search is rather inefficient, but it is primitive recursive and so we can express it as a bounded formula.

There is a similar formula $accept_{time}(k, m, n)$ in which the running time is bound by $|n|^m$.

Now consider the formula:
$$\forall k, m . \exists k', m' . \forall n .
accept_{space}(k,m,n) \Leftrightarrow accept_{time}(k',m', n).
$$
It says that for every machine $k$ which uses at most space $|n|^m$ there is a machine $k'$ which uses at most time $|n|^{m'}$ such that the two machines accepts exactly the same $n$'s. In other words, the formula says $PSPACE \subseteq P$. This formula is $\Pi^0_3$.

We can improve this if we are willing to express instead the sentence "$TQBF$ is in polytime" , which should be good enough for most applications, as TQBF is PSPACE complete and so it being in polytime is equivalent to $PSPACE \subseteq P$. Let $k_0$ be (the code of) a machine which recognizes TQBF in space $|n|^{m_0}$. Then "$TQBF \in P$" can be expressed as
$$\exists k', m' . \forall n . accept_{space}(k_0, m_0, n) \Leftrightarrow accept_{time}(k', m', n).$$
This formula is just $\Sigma^0_2$. If I were a complexity theorist I would know if it is possible to do even better (but I doubt it).