Does the Kolmogorov complexity of a program $p$ generating a string $x$ equal the complexity of $x$ up to constant?
I guess, I have found the answer myself: no.

Let $q$ be a shortest program that generates $x$: $K(x)=l(q)$. It is enough to concatenate any incompressible string $s$ not sharing information with $q$ and setting $p=\bar{q}s$. Let $m$ be the number of a Turing machine $T_m$ in the standard enumeration, that takes $p$, executes $q$ and discards $s$. Then, $U(\langle m,p\rangle)=T_m(p)=x$. 
However, $K(p)\ge l(q)+l(s)+O(1)&gt;l(q)=K(x)$. We thus have constructed a $p$ that contradicts the claim.