Q:

Concatenation preprocessing:

This line :
#define hash_hash # ## #

will produce ## but isn't that Undefined Behavior based on :

If the result is not a valid preprocessing token, the behavior is
  undefined.

EDIT:

In other words, expanding hash_hash produces a new token, consisting
  of two adjacent sharp signs, but this new token is not the ##
  operator.

The categories of preprocessing tokens are:

Preprocessing tokens fall into five broad classes: identifiers,
  preprocessing numbers, string literals, punctuators, and other

Is ## new token falls in other category if there is no UB. 
Note: please take your times to read the question and answer according to it or even before downvoting it.

A:

That is an example from the standard. ## is a valid preprocessing token. It is the token which forms the concatenation operator, as used right there. It simply has no meaning in the C grammar (after preprocessing). But it appears in the list of punctuation tokens, equally as * or ++.
Since no grammar rule will match it, the inevitable result is a syntax error which will be diagnosed. Unless, it is passed to another macro which stringizes or ignores it.
#define IGNORE(X)
#define STRINGIZE_LIT(X) #X
#define STRINGIZE(X) STRINGIZE_LIT(X)

IGNORE( hash_hash ) // OK, token is not passed to semantic analysis.
printf( STRINGIZE( hash_hash ) ); // OK, same as printf( "##" );

