No, the different $p$-adic number systems are not in any way compatible with one another.
A $p$-adic number is a not a number that is $p$-adic; it is a $p$-adic number. Similarly, a real number is not a number that is real, it is a real number. There is not some unified notion of "number" that these are all subsets of; they are entirely separate things, though there may be ways of identifying bits of them in some cases (e.g., all of them contain a copy of the rational numbers).
Now, someone here is bound to point out that if we take the algebraic closure of some $\mathbb{Q}_p$, the result will be algebraically isomorphic to $\mathbb{C}$. But when we talk about $p$-adic numbers we are not just talking about their algebra, but also their absolute value, or at least their topology; and once you account for this they are truly different. (And even if you just want algebraic isomorphism, this requires the axiom of choice; you can't actually identify a specific isomorphism, and there's certainly not any natural way to do so.)
How can we see that they are truly different? Well, first let's look at the algebra. The $5$-adics, for instance, contain a square root of $-1$, while the $3$-adics do not. So if you write down a $5$-adic number which squares to $-1$, there cannot be any corresponding $3$-adic number.
But above I claimed something stronger -- that once you account for the topology, there is no way to piece the various $p$-adic number systems together, which the above does not rule out. How can we see this? Well, let's look at the topology when we look at the rational numbers, the various $p$-adic topologies on $\mathbb{Q}$. These topologies are not only distinct -- any finite set of them is independent, meaning that if we let $\mathbb{Q}_i$ be $\mathbb{Q}$ with the $i$'th topology we're considering, then the diagonal is dense in $\mathbb{Q}_1 \times \ldots \times \mathbb{Q}_n$.
Put another way -- since these topologies all come from metrics -- this means that for any $c_1,\ldots,c_n\in\mathbb{Q}$, there exists a sequence of rational numbers $a_1,a_2,\ldots$ such that in topology number 1, this converges to $c_1$, but in topology number two, it converges to $c_2$, and so forth. (In fact, more generally, given any finite set of inequivalent absolute values on a field, the resulting topologies will be independent.)
So even on $\mathbb{Q}$, the different topologies utterly fail to match up, so there is no way they can be pieced together by passing to some larger setting.
Imagine that $K=\mathbb{Q}[\alpha]$ for some algebraic number $\alpha$. Let $m(x)$ be the minimal polynomial of $\alpha$. Then we know that $K\cong \mathbb{Q}[x]/\langle m(x)\rangle$. As $\mathbb{Q}_p$ is an extension field of the rationals, the polynomial $m(x)$ (irreducible in $\mathbb{Q}[x]$) may factor into a product of irreducible factors
$$
m(x)=\prod_i m_i(x),
$$
where $m_i(x)$ are irreducible polynomials in $\mathbb{Q}_p[x]$. Because $m(x)$ is separable, the polynomials $m_i(x)$ are distinct. Therefore the Chinese Remainder Theorem says that
$$
\mathbb{Q}_p[x]/\langle m(x)\rangle\cong\bigoplus_i \mathbb{Q}_p[x]/\langle m_i(x)\rangle.
$$
Here on the right hand side the summands $\mathbb{Q}_p[x]/\langle m_i(x)\rangle$ are all extension fields of $\mathbb{Q}_p$, because the polynomials $m_i(x)$ are irreducible. OTOH
$$
\mathbb{Q}_p[x]/\langle m(x)\rangle\cong\mathbb{Q}[x]/\langle m(x)\rangle\otimes\mathbb{Q}_p\cong K\otimes\mathbb{Q}_p.
$$
Undoubtedly you can guess the rest - the fields $\mathbb{Q}_p[x]/\langle m_i(x)\rangle$ are the fields $K_{p,i}$.
Best Answer
I don't know about the direct (inductive) limit, but the definition of $\mathbf Z_p$ is that it is the inverse or projective limit: $$\mathbf Z_p=\varprojlim \mathbf Z/p^n \mathbf Z=\Bigl\{(\bar x_n)\in\prod_n\mathbf Z/p^n \mathbf Z\mid x_{n+1}\equiv x_n\bmod p^n \Bigr\}$$