First question:
Let's prove the contrapositive. Assume that for all $n$ and all $L(A)$-formulas $\varphi(x)$, $p(x)\cup \{\varphi(x)\}\not\models p_n(x)$. We want to prove that $p(x)\not\models \bigvee_{n\in\omega} p_n(x)$, i.e. we want to realize $p(x)$ by an element which does not satisfy any of the $p_n(x)$.
One way to go is to introduce a new constant symbol $c$, consider the theory $T\cup p(c)$, and use the omitting types theorem to omit all the $p_n(x)$. But this is a little messy, what with the changing of languages, and it's also overkill: you don't really have to omit all the types $p_n(x)$, just make sure that $c$ doesn't realize them.
So I would prefer to prove this result by redoing a small part of the proof of the omitting types theorem. Note that we don't have to assume the language is countable in this proof, which is a big advantage over using omitting types!
We build a sequence of $L(A)$-formulas $(\varphi_n(x))_{n\in \omega}$ by induction, ensuring as we go that for all $n$, $p(x)\cup \{\varphi_k(x)\mid k\leq n\}$ is consistent. To pick $\varphi_n(x)$, we let $\psi_n(x) = \bigwedge_{k<n}\varphi_k(x)$. By induction, $\psi_n(x)$ is consistent with $p(x)$. So by our assumption, $p(x)\cup \{\psi_n(x)\}\not\models p_n(x)$. Thus we can find some element $b$ satisfying $p(x)\cup \{\psi_n(x)\}$ but not satisfying $p_n(x)$. Let $\varphi_n(x)$ be some formula which is true of $b$, but whose negation is in $p_n(x)$. Then $p(x)\cup \{\varphi_k(x)\mid k\leq n\}$ is consistent, since it is realized by $b$.
Now let $q(x) = p(x)\cup \{\varphi_n(x)\mid n\in \omega\}$. By compactness, $q(x)$ is consistent. Let $c$ be some realization. Then $c$ realizes $p$, but $c$ does not realize any $p_n(x)$, since for any $n$, $c$ satisfies $\varphi_n(x)$, and $\lnot\varphi_n(x)\in p_n(x)$.
Second question:
To me, this question seems totally unrelated to the first question and the omitting types theorem... I would prove it like this:
Let $Q$ be the set of complete types $q(x)$ over $B$ with $p(x)\subseteq q(x)$. For each $q(x)\in Q$, $q(x)$ is isolated by an $L(B)$-formula $\varphi_q(x)$. So $p(x)\cup \{\lnot \varphi_q(x)\mid q\in Q\}$ is inconsistent. By compactness, there are finitely many types $q_1,\dots,q_n$ such that $p(x)\models \bigvee_{i=1}^n \varphi_{q_i}(x)$. Let's call this disjunction $\psi(x)$. Note that also $\psi(x)\models p(x)$, since if $\psi(a)$, then $a$ realizes $q_i(x)$ for some $i$, and $p(x)\subseteq q_i(x)$. Thus we have shown that $p(x)$ is isolated by a formula over $B$.
The question as stated is a bit ambiguous, but I suppose the goal is to show that $p(x)$ is isolated by a formula over $A$. But this follows from the above, since if a set definable over $B$ is invariant over $A\subseteq B$ (i.e. fixed by all automorphisms of the monster model which fix $A$ pointwise), then the set is actually definable over $A$. I'll leave that as an exercise for you.
(1) "Formulas in different types should be inconsistent". This is wrong. If $p$ and $q$ are distinct types, then there is some formula $\varphi$ such that $\varphi\in p$ and $\lnot \varphi\in q$, and clearly $\varphi$ and $\lnot \varphi$ are inconsistent. This does not mean that for any $\varphi\in p$ and any $\psi\in q$, $\varphi$ and $\psi$ are inconsistent. For example, the valid formulas like $x = x$ will be in every type.
(2) Every formula $\varphi$ determines the set of types $[\varphi] = \{p\mid \varphi\in p\}$. If $[\varphi] = [\psi]$, then $\varphi$ and $\psi$ are equivalent. (Why? if $\varphi$ and $\psi$ were not equivalent, there would be some element $a$ in some model such that, without loss of generality, $a$ satisfies $\varphi$ and $\lnot \psi$. Then $\mathrm{tp}(a)\in [\varphi]$ and $\mathrm{tp}(a)\notin [\psi]$, so $[\varphi]\neq [\psi]$.) So if there are $2^m$ sets of types, there are at most $2^m$ sets of the form $[\varphi]$, so there are at most $2^m$ formulas up to equivalence.
Best Answer
The answer to your first question is a strong no whenever $T$ has infinite models. Indeed, if $M$ is any infinite model of $T$, then there is a consistent type $p(v)\in S_1(M)$ containing the set of formulas $\{v\neq m:m\in M\}$. This type cannot be isolated, since any isolated type is realized, and $p(v)$ is not realized in $M$. If you prefer an example where the type in question is realized in $M$, here's a different one; let $T=\operatorname{DLO}$ and let $M=\mathbb{Q}$ be the standard model, and take $A$ to be the subset $\{1/n:n\in\mathbb{N}\}$. Then $\operatorname{tp}(0/A)$ is not isolated, since for any finite subset $A_0\subset A$ there exists $c>0$ with the same type as $0$ over $A_0$, but there does not exist any $c>0$ with the same type as $0$ over $A$.
However, if $A$ is finite, then the answer is yes. Indeed, in that case, enumerate $A$ as $a_1\dots a_m$, and let $p(v_1,\dots,v_n)$ be a type in $S_n(A)$. Let $q(v_1,\dots,v_n,w_1,\dots,w_m)\in S_{m+n}(\varnothing)$ be the type obtained by replacing every instance of $a_i$ in a formula of $p$ by the variable $w_i$. By Ryll-Nardzewski, $q$ is isolated, say by a formula $\phi(\overline{v},\overline{w})$. Then $p$ is isolated by $\phi(\overline{v},\overline{a})$ (why?), as desired.
Finally, the answer to your second question is yes; taking $A=\varnothing$ gives precisely the Ryll-Nardzewski criterion that you cite at the beginning of your answer.