It may be the case that no prime $v$ of $K$ splits completely in $L$. Indeed, that would be the "typical" outcome.

If there exists at least one non-cyclotomic $\mathbf{Z}_p$-extension of $K$, there exists a $\mathbf{Z}^2_p$-extension $M/K$ which contains the cyclotomic extension. Let $\Gamma = \mathrm{Gal}(M/K)$. Suppose that $v \nmid p$. Because $v$ does not split completely in the cyclotomic extension, the decomposition group $D_v \subset \Gamma$ is isomorphic to $\mathbf{Z}_p$ (It is cyclic and non-trivial). A choice of $\mathbf{Z}_p$ extension corresponds to a choice of saturated subgroup

$$\mathbf{Z}_p \simeq H \subset \Gamma,$$

where $\mathrm{Gal}(L/K) = \Gamma/H$.

The prime $v$ splits completely in $L/K$ if and only if $H$ contains the subgroup $D_v \simeq \mathbf{Z}_p
$ of $\Gamma$. Now there are uncoutably many choices of subgroup $H$ (and hence choices of $L$) and only finitely many contain any fixed subgroup $D_v$, and so only countably many contain a $D_v$ as $v$ ranges over all primes in $K$. Hence, for a "generic" $\mathbf{Z}_p$ extension, no single prime $v$ will split completely. (It's easy to incorporate into this argument the primes $v|p$ as well.)

I'm not sure if this answers your question or not