Computability and Forcing In view of the paper "Forcing As A Computational Process" by J. Hamkins, R. Miller and K. Williams, I have revised my original question, part (a) about the computability of the Forcing Truth Definition, Looking into the Future within Forcing as follows (instead of Cohen's book, using Shoenfield paper :
https://mathweb.ucsd.edu/~sbuss/CourseWeb/Math260_2012F2013W/Shoenfield_UnramifiedForcing.pdf)
In demonstrating the definability of forcing in the base model M, the definition of $\Vdash$* includes (page 362 definition (c)):
p $\Vdash$* $\neg$ $\Phi$ $\;$ iff $\;$  For all q $\leqq$ p $\neg$ (q $\Vdash$* $\Phi$ )  .....................(1)
Although the complexity of the construction of a formula $\Phi$ is finite, there is an infinite number of finite q $\leqq$ p to test for q $\Vdash$* $\Phi$. If none of the q lead to q $\Vdash$* $\Phi$, this would mean that a completed infinity of tests may be required to establish (1) making the overall calculation not computable.
However "Forcing As A Computational Process" in Main Theorem 1 on page 1 says that : G is computable from the Atomic Diagram of M.
So I was wondering whether anyone could help to identify the error in thinking that "determination of expressions like (1) could require a completed infinity of calculations."
 A: This is set theory, not computability theory.  Presumably Shoenfield doesn't claim that $p\Vdash^*\Phi(t_1, t_2, \dots)$ is computable, just that it's definable (in ZF or ZFC, whichever he's taking as the base theory).  To be precise, this definability is provable in set theory for a fixed formula $\Phi,$ or for all $\Sigma_n$ formulas $\Phi$ for a fixed $n.$  (You can't prove the definability of $p\Vdash^*\Phi(t_1, t_2, \dots)$ with $\Phi$ as a parameter varying over all formulas, because you'd run afoul of Tarski's undefinability of truth.)

If the question is how to reconcile this with $\Vdash^*$ being computable from information about the ground model, that statement is true, but it's a relative computability fact, relative to a collection of facts about the ground model (which is infintely many facts).  This doesn't mean that $\Vdash^*$ is actually computable.
Addendum: I just glanced at the paper by Hamkins et al. and they talk about computing some $G$ or $M[G]$ relative to the appropriate first-order facts about $M.$ But $G$ and $M[G]$ aren’t computable outright; it’s all relative to $M$ (and even this relative computability is only for some generic filters).
A: Note that in the Forcing as a computational process paper, the theorem merely states that some generic is computable from (the atomic diagram of) $M$, not that every generic is.
Proof:

The proof of the theorem is roughly this: from $M$, we can decide
whether any given $p\in M$ is in $\mathbb{P}\in M$, and similarly
whether or not $p\leqslant^{\mathbb{P}}q$ for $p,q\in \mathbb{P}$.
Since $M$ has the set of all its dense subsets of $\mathbb{P}$ (and
the set of all non-dense subsets), it's also decidable from $M$
whether or not any given subset of $\mathbb{P}$ is dense or not.  So
$\mathbb{P}$, $\leqslant^{\mathbb{P}}$, being dense, and so on are all
decidable relative to $M$.  How do we therefore compute a new generic?
Well the same way we would for generating a generic from any countable
transitive model: just continually extend $q_n$ into the next dense
set $D_{n+1}$ to get $q_{n+1}$. The upward closure of this $\{q_n:n\in
 \omega\}$ is the resulting generic $G$.  Note that
$\{q_n:n\in\omega\}$ is decidable from $M$.
So to compute membership $p\in G$, we just need $p$ to be above the
"least" element $q_n\in D_n$ such that $q_n$ extends $q_m$ for all
$m<n$, and where $D_n$ in the $n$th dense set.  This gives an
algorithm that outputs "yes" if $p\in G$, basically meaning $G$ is
$\Sigma^0_1$ relative to $M$.  To determine if $p\not\in G$, we use
antichains: if we consider two incompatible elements $p,q\in
 \mathbb{P}$, if $p\not\in G$ then $p$ must be incompatible with some
element of $G$ (otherwise just take the dense set of things below $p$
in union with things incompatible with $p$ and see which $G$ can
intersect).  As a result, to check whether $p\not\in G$, we just need
to check if the pair $\langle p,q_n\rangle$ is in the set of
incompatible pairs of elements of $\mathbb{P}$ for some $n\in\omega$
where $q_n$ is as above.  This gives an algorithm that spits out "no"
to the question "is $p\in G$?" whenever $p\not\in G$.  In other words,
$G$ is $\Pi^0_1$ relative to $M$ and hence is $\Delta^0_1$ (i.e.
decidable) relative to $M$.$\blacksquare$

Now what does this tell us about the computability of the forcing relation?  Not much.  The paper goes into this a bit, talking about how with only access to the membership relation in $M$, we can't compute much about the theory of $M[G]$ (where $G$ is computable from $M$ as above).  With access to the entire theory of $M$, we can compute more of $M[G]$, as expected, with negation increasing the complexity needed to compute more about $M[G]$.
