How can we prove a statement is provable? Given a concrete mathematical statement, such as BSD conjecture(https://en.wikipedia.org/wiki/Birch_and_Swinnerton-Dyer_conjecture), do we know if it is provable?
 A: You're using the wrong term. You mean to ask whether we can tell if a conjecture is decidable, meaning that it is either provable or disprovable. But no we cannot tell whether a statement is decidable if the quantifier complexity is too high. Furthermore, it may be possible that even the decidability of a statement is itself undecidable! (See below for an example.)
First read https://math.stackexchange.com/a/1643073/21820, to ensure that you fully understand the import of Godel's incompleteness theorem. After that, consider the following.
$\def\imp{\rightarrow}$
[We work in a meta-system and assume that $PA$ is omega-consistent.]
Let $φ = \square_{PA} Con(PA) \lor \square_{PA} \neg Con(PA)$. [So $φ$ expresses "Con(PA) is decidable over $PA$".]
If $PA \vdash φ$:
  Within $PA$:
    $\square Con(PA) \lor \square \neg Con(PA)$.
    If $\square Con(PA)$:
      $\neg Con(PA)$. [by the internal incompleteness theorem]
      $\square \bot$.
      $\square \neg Con(PA)$. [by (D1),(D2)]
    $\square \neg Con(PA)$. [by basic logic]
    $\neg Con(PA)$. [because $PA$ is omega-consistent]
  Contradiction. [with the external incompleteness theorem]
Therefore $PA \nvdash φ$.
If $PA \vdash \neg φ$:
  Within $PA$:
    $\neg \square Con(PA)$. [by basic logic]
    If $\square \bot$:
      $\square Con(PA)$. [by (D1),(D2)]
      Contradiction.
    $\neg \square \bot$.
    $Con(PA)$.
  Contradiction. [with the external incompleteness theorem]
Therefore $PA \nvdash \neg φ$.
Thus $φ$ is independent of $PA$.
