For a givien partial order, how many generic extensions might exist? In other words, for a boolean valued model class which dreams of a generic extension, how many unique generic objects exist for a given partial order (which gives rise to a boolean algebra naming such an extension)?

It is often the case that when we force with a partial order $\mathbb{P}$, then in the resulting forcing extension $V[G]$ there may be other $V$generic filters and indeed many other $V$generic filters $G'$, even giving rise to the same extension $V[G]=V[G']$. Thus, although from a the filter $G$ we may construct the forcing extension $V[G]$, we should not think of the of the forcing extension $V[G]$ as necessarily determining $G$; different filters can lead to the same forcing extension. For example, if $\mathbb{P}$ has an automorphism $\pi:\mathbb{P}\to\mathbb{P}$ in $V$, then it is not difficult to see that $\pi"G$ is also $V$generic, and furthermore these generic filters give rise to the same forcing extension $V[G]=V[\pi"G]$. Since many of the mostcommonlyused forcing notions admit numerous automorphisms, it is quite common that we find ourselves in a forcing extension $V[G]$ for which there are numerous filters $G'\in V[G]$ for the same forcing notion $\mathbb{P}$ with $V[G]=V[G']$. And we can often count them in a precise way, as Stefan did in his answer. In particular, highly homogeneous forcing notions, having numerous automorphisms, will always lead to forcing extensions having many $V$generic filters. Meanwhile, there are also interesting cases where the forcing extension necessarily has fewer or sometimes only one generic filter for that order, which it seems is part of what you are asking about. For example, from $\Diamond$ one can construct Souslin trees with the unique branch property, meaning that the tree in the forcing extension has precisely one branch (and with a Souslin tree, the generic filters in any extension are precisely the branches), and one can view this property of a forcing notion as a strong kind of rigidity. This kind of situation with Souslin trees was the theme of my article with Gunter Fuchs on the Degrees of rigidity for Souslin trees, where we separate various strong notions of rigidity into a hierarchy. Another example is selfencoding forcing, the forcing notion that adds a subset of some cardinal $\kappa$, and then codes this set into the GCH pattern (or whatever), and then codes these new sets into the GCH pattern a bit higher up, and so on. The process catches its tail in $\omega$ many steps, and the result is that in the corresponding forcing extension $V[G]$, there is only one $V$generic filter for the poset, and it is therefore definable there from parameter $\kappa$. When a nontrivial forcing notion $\mathbb{P}$ necessarily leads to a forcing extension in which $\mathbb{P}$ has exactly one generic filter, then one should think of $\mathbb{P}$ as being far from homogeneous. 


It is a bit difficult to understand the question. But are you asking for the number of different forcing extensions by a given partial order? From the perspective of the universe that we are living in? Or from the outside? Let me be more specific. Let $M$ be a countable transitive model of set theory.
In $M$, let $P$ be Cohen forcing, i.e., the partial order of functions from a natural number
to $2$, ordered by reverse inclusion. In the real world, there are $2^{\aleph_0}$ different $P$generic filters over $M$. Now, if you are interested in the number of extensions from the perspective of $M$,
that question does not really make sense, since $M$ doesn't have access to the generic filters over $M$. Since $\kappa$ was an arbitrary cardinal of $M$, this shows that it is impossible to gauge the number of generic extensions from inside the ground model. I hope this does at least tell you how the question should be phrased. 


Let me add an easier observation about the number of $M$generic filters, for a given poset $P\in M$, if we do the counting "from the outside". That is, suppose $M$ is a countable transitive model and we want the number of generic filters in the real world (which, in particular, sees a counting of $M$). As Stefan pointed out, there is an obvious upper bound of $2^{\aleph_0}$ because there are only that many subsets of $P$ altogether. On the other hand, this bound is attained by any nontrivial forcing $P$, where "nontrivial" means that there is some $q\in P$ such that every extension of $q$ has at least two incompatible extensions. (If this fails, then the generic extensions are just the ground model.) In this situation, the usual proof that generic filters exist (namely, list $M$'s dense subsets of $P$ in an $\omega$sequence and build the generic filter in $\omega$ steps, hitting the $n$th dense set at the $n$th step) is easily modified to give $2^{\aleph_0}$ different generic filters, by inserting steps at which you choose arbitrarily one of two (previously selected) incompatible extensions of the condition chosen at the previous step. As Stefan pointed out, once you have $2^{\aleph_0}$ generic filters, you also have $2^{\aleph_0}$ generic extensions, because each of the extensions contains at most countably many of the generic filters. 

