NP complete problems related to permutations of binary vectors or block ciphers
In a strict sense, no.  NP is about worst-case hardness.  Cryptography requires average-case hardness.  $P \ne NP$ implies the existence of problems that are hard in the worst-case (the worst-case running time is super-polynomial) but says nothing about average-case hardness.

For block ciphers, we need average-case hardness.  Therefore, there are good reasons to expect that $P\ne NP$ won't be enough.

(There are other reasons to think that $P\ne NP$ is not enough -- see, e.g., Impagliazzo's many worlds of cryptography -- but I'm trying to keep it simple for you.)

In a loose sense, you might be able to find some loose relationship, e.g., where someone builds a block cipher whose security they claim is "based" upon the NP-completeness of some problem.  But that kind of claim is on very shaky ground.  It's very unlikely they will have proven a theorem that says "if $P\ne NP$, then my block cipher is secure".  Instead, they will have some handwavy argument about how the problem is "related", and it'll all be just a bit sketchy.



The blog post you mention actually describes a relationship that goes in the opposite way: if secure ciphers exist, then a large class of natural ways of trying to prove $P\ne NP$ are doomed to fail.  That's a pretty nifty result.