Let $A$ and $B$ be two $C^*$algebras, and let $p:A \to B$ be a surjective normdecreasing $*$homomorphism which is injective on a dense $*$subalgebra of $A$. Can such a map have nontrivial kernel, and if so, is it possible that the $K$theory groups of $A$ and $B$ can be nonisomorphic?

2$\begingroup$ All starHMs between Cstar algebras are automatically normdecreasing (assuming you mean what I would call contractive) $\endgroup$– Yemon ChoiMar 4 '20 at 23:54
Yes to both.$\newcommand{\Cst}{{\rm C}^*}$ The standard example for the first is: take a discrete group $G$ and let $A$ be its full $\Cst$algebra, $B$ its reduced $\Cst$algebra. There is a canonical homomorphism $q:A\to B$ which is injective when restricted to $\ell^1(G)$; but $q$ is injective if and only if $G$ is amenable. So any nonamenable discrete group will provides examples for your first question.
There are nonamenable groups $G$ for which $q$ induces an isomorphism on $K$theory — I think the standard name for such groups is $K$amenable. Lance proved that free groups have this property. But infinite groups with Kazhdan's Property (T) do not have this property because the socalled Kazhdan projection in the full group $\Cst$algebra lies in the kernel of $q$. (Thanks to Jamie Gabe in comments for clarifying/sharpening my original statement.)
(See also this MO question $*$algebras, completions, and $K$theory )

4$\begingroup$ It might be worth adding that the Kazhdan projection $p$ has nontrivial $K_0$class in the kernel of $q$ since $p$ is mapped to a generating projection in $K_0(\mathbb C) = \mathbb Z$ by the trivial representation. $\endgroup$ Mar 5 '20 at 0:14
There are even commutative counterexamples. Let $A = C[0,2]$ and let $A_0$ be the $*$subalgebra of all polynomials in $x$. Then let $p: C[0,2] \to C[0,1]$ be the restriction map.
(My first example took $A = C[0,3]$ and let $A_0$ be the set of all polynomials in $x$ with rational coefficients and $p: A \mapsto \mathbb{C}$ the point evaluation at $x = e$. Since $e$ is transcendental, $p$ is injective on $A_0$. But $A_0$ is not a $*$subalgebra over $\mathbb{C}$.)