I keep all your notations but I'll introduce also the $\Lambda$-modules $X$ (rep. $X'$) = Galois group over $K$ of the maximal abelian pro-$p$-extension of $K$ which is uramified (resp. is unramified and splits totally all the $p$-places of $K$). It is known that the Pontryagin dual of $C(p)$ is pseudo-isomorphic to $X^*$ = the module $X$ with inverted action of $\Gamma = \bar{<\gamma>}$. The same property holds for $X'$. It follows that the finiteness of $C(p)^{\Gamma}$ is equivalent to that of $X^*_{\Gamma}$, i.e. $T^*=\gamma^{-1} -1$ does not divide the characteristic series of $X^*$, or equivalently, ${X^*}^{\Gamma}$ is finite, or equivalently, ${X}^{\Gamma}$ is finite. So we are brought back to construct an example of an infinite ${X}^{\Gamma}$.
Suppose moreover that $k$ is CM and $p$ is odd. Let $r$ be the number of $p$-places of $k^+$ which split completely in $k$. It is known that the kernel of the canonical surjection $X^- \to (X')^-$ is isomorphic to a direct sum $r$ factors of the form $\Lambda/\omega_{n_i}\Lambda$, where $\omega_n = {\gamma}^{p^n}-1$ (a more precise statement is in the appendix of [FG]). This implies that $(X^-)^{\Gamma}$ is infinite, and we are done.
[FG] L.-J. Federer, B.-H. Gross: Regulators and Iwasawa modules, Invent. Math, 62 (1981), 443-457
Let $f(x,y) = ax^3 + bx^2y + cxy^2 + dy^3$ for $a,b,c,d \in \mathbf Q$. If $a$ or $d$ were $0$ then $f(1,0) = a$ would be $0$ or $f(0,1) = d$ would be $0$. Therefore the assumption that $f$ has no zeros in $\mathbf P^1(\mathbf Q)$ implies $a$ and $d$ are not $0$.
Now consider
$$
F(x) = f(x,1) = ax^3 + bx^2 + cx + d,
$$
which is cubic in $\mathbf Q[x]$ with no rational root: if $F(r) = 0$ for some $r \in \mathbf Q$ then $f(r,1) = 0$, but $f$ has no rational zero. A cubic in $\mathbf Q[x]$ with no root in $\mathbf Q$ is irreducible over $\mathbf Q$ (an accident of small-degree polynomials!), so $F(x)$ is irreducible over $\mathbf Q$. The following theorem explains why $F(x)$ has no root in some $\mathbf Q_p$.
Theorem. If $F(x) \in \mathbf Q[x]$ is irreducible over $\mathbf Q$ and has a root in $\mathbf Q_p$ for all $p$ then $F(x)$ is linear.
The same result and proof hold with $\mathbf Q$ replaced by an arbitrary number field (and the $\mathbf Q_p$'s replaced by all the nonarchimedean completions of that number field), but I stick to coefficients in $\mathbf Q$ to keep the notation simple.
Proof. Let $K = \mathbf Q(\alpha)$ where $F(\alpha) = 0$ and let $L$ be the Galois closure of $K$ over $\mathbf Q$.
The discriminant ${\rm disc}(F)$ is nonzero since $F(x)$ is irreducible over $\mathbf Q$.
Let $S$ be the primes $p$ that appear in the factorization of ${\rm disc}(F)$ or that ramify in $L$ or that appear in a coefficient of $F(x)$.
That is a finite set of primes. For $p \not\in S$, $F(x)$ is $p$-integral and $F(x) \bmod p$ is separable since ${\rm disc}(F) \not\equiv 0 \bmod p$.
For $p \not\in S$, $F(x)$ has a root $r$ in $\mathbf Q_p$ by hypothesis, and the root is in $\mathbf Z_p$ since $F(x)$ has $p$-integral coefficients and its leading coefficient (in fact each coefficient) is a $p$-adic unit.
Therefore we can reduce the equation $F(r) = 0$ to $\overline{F}(\overline{r}) = 0$ in $\mathbf Z_p/(p) = \mathbf F_p$. We have $K \cong \mathbf Q(r)$, so some prime ideal factor of $p$ in $K$ has residue field degree $1$: the embedding of $K$ into $\mathbf Q_p$ by mapping $\alpha$ to $r$ puts a non-archimedean absolute value $|\cdot|_{\mathfrak p}$ on $K$ with residue field $\mathbf F_p$, so $f(\mathfrak p|p) = 1$.
Now we bring in Frobenius elements. Set $G = {\rm Gal}(L/\mathbf Q)$ and $H = {\rm Gal}(L/K)$. Since $p$ is unramified in $L$ (by $p \not\in S$), we can talk about a Frobenius element at a prime $\mathfrak P$ over $p$ in $L$. Since $F(x)$ is $p$-integral and $F(x) \bmod p$ is separable, the permutation action of ${\rm Frob}(\mathfrak P|p)$ on the roots of $F(x)$ has a cycle structure that matches the degrees of the irreducible factors of $F(x) \bmod p$. (The actual choice of $\mathfrak P$ over $p$ doesn't matter, since all Frobenius elements at primes over $p$ are conjugate to each other in $G$ and thus permute the roots of $F(x)$ with the same cycle structure.)
We know $F(x) \bmod p$ has a root in $\mathbf F_p$, so it has a linear irreducible factor mod $p$. Thus ${\rm Frob}(\mathfrak P|p)$ has a fixed point in its permutation action on the roots of $F(x)$: this Frobenius element fixes a root of $F(x)$ in $L$, so this Frobenius element fixes one of the subfields of $L$ conjugate to $K = \mathbf Q(\alpha)$. The subgroup of $G$ corresponding to that subfield is conjugate to $H$. Hence ${\rm Frob}(\mathfrak P|p) \in \bigcup_{\sigma \in G} \sigma H\sigma^{-1}$.
Now we bring in the Chebotarev density theorem: every element of $G$ is a Frobenius element at infinitely many primes, in fact at a positive density of primes. So each $g \in G$ is a Frobenius element at a prime ideal lying over a prime outside of $S$. The above argument proves $g \in \bigcup_{\sigma \in G} \sigma H\sigma^{-1}$. Thus
$$
G = \bigcup_{\sigma \in G} \sigma H\sigma^{-1}.
$$
A finite group is never the union of conjugates of a proper subgroup, so $H = G$ and thus $K = \mathbf Q$. Therefore $\mathbf Q(\alpha) = \mathbf Q$, so $F(x)$ has a root in $\mathbf Q$. Since $F(x)$ is irreducible over $\mathbf Q$ with a rational root, $F(x)$ is linear. QED
In this proof, we did not need $F(x)$ to have a root in $\mathbf Q_p$ at every prime $p$. It is okay to say $F(x)$ has a root in $\mathbf Q_p$ outside a set $S$ of primes of density $0$, since Chebotarev implies that each $g \in G$ is a Frobenius at a prime outside $S$. For example, if $F(x)$ has a root in $\mathbf Q_p$ for all but perhaps finitely many $p$ then we can still deduce that $F(x)$ has a root in $\mathbf Q$.
Therefore an irreducible $F(x) \in \mathbf Q[x]$ of degree greater than $1$ has no root in $\mathbf Q_p$ for infinitely many $p$. When $F(x)$ is cubic, not having a root in a field is equivalent to being irreducible over that field. Therefore a cubic irreducible in $\mathbf Q[x]$ has no root in $\mathbf Q_p$ for infinitely many $p$. Returning to the binary cubic form, $f(x,1)$ has no root in $\mathbf Q_p$ of infinitely many $p$, so for infinitely many $p$ we can't solve $f(x,y) = 0$ with $[x,y] \in \mathbf P^1(\mathbf Q_p)$ with $y \not= 0$. Of course we can't solve it with $y = 0$ either, since that would be the point $[x,0] = [1,0]$ and $f(1,0) = a \not= 0$.
Best Answer
Suppose that $K$ is ramified at $p_1,\ldots,p_n$. Then certainly $K \subset \mathbb Q(\zeta_{p_1^{m_1}},\ldots,\zeta_{p_n^{m_n}})$ for some $m_1,\ldots,m_n$.
Now you could try to pin down the field precisely by using some combination of information about the discriminant and the Galois group of $K/\mathbb Q$.
For example, suppose that our abelian extension $K$ is ramified at a single odd prime $p$. Then $Gal(\mathbb Q_{\zeta_{p^m}})/\mathbb Q) = (\mathbb Z/p^m\mathbb Z)^{\times}$ is cyclic; in fact it is a product of the cyclic groups $(1 + p\mathbb Z_p)/(1+p^m\mathbb Z_p)$ (of order $p^{m-1}$) and $(\mathbb Z/p\mathbb Z)^{\times}$ (of order $p-1$).
Any quotient is then equal to $(1+p\mathbb Z_p)/(1+p^{m'}\mathbb Z_p)$ (for $1 \leq m'\leq m$) and some quotient $H$ of $(\mathbb Z/p\mathbb Z)^{\times}$.
Thus if we choose $m$ minimally in the first place, the Galois group of $K$ over $\mathbb Q$ (which is a quotient of $Gal(\mathbb Q_{\zeta_{p^m}}/ \mathbb Q)$) is isomorphic to $(1 + p\mathbb Z_p)/(1+p^m\mathbb Z_p)\times H$, for a quotient $H$ as above.
The problem that you asked about is to compute $m$ given $K$. In this case we see that it is just a matter of computing the power of $p$ dividing the order of $Gal(K/\mathbb Q)$: the number $m$ is one more than the power of $p$ dividing this order.
As a slightly more complicated example, suppose that $K$ is ramified at two odd primes $p < q$, so that $K \subset \mathbb Q(\zeta_{p^m},\zeta_{q^n})$. Again, let's choose $m$ and $n$ minimally. Then $Gal(K/\mathbb Q)$ is a quotient of $$(\mathbb Z/p^m\mathbb Z)^{\times} \times (\mathbb Z/q^n\mathbb Z)^{\times}$$ $$ = (\mathbb Z/p\mathbb Z)^{\times} \times (1 + p\mathbb Z_p)/(1+p^m\mathbb Z_p) \times (\mathbb Z/q\mathbb Z)^{\times} \times (1+q\mathbb Z_q)/(1+q^n\mathbb Z_q)^{\times} .$$
Now we see that it makes a difference whether or not $p | q - 1$. More precisely, suppose that $p^e$ is the precise power of $p$ that divides $q-1$. Then (thinking about the possible Sylow subgroup structures of a quotient of the above product, and the fact that $m$ and $n$ were chosen minimally), we see that $n$ is one more than the power of $q$ dividing the order of $Gal(K/\mathbb Q)$. On the other hand, if $m'$ is the power of $p$ dividing this order, then what we see is that $m-1 \leq m' \leq m-1+e$, so if $e \geq 1$, then we have not pinned down $m$ precisely just by knowing $m'$.
However, if we now apply the conductor-discriminant formula, one will find that knowledge of the discriminant of $K$ should allow us to determine $m$. The point is that the contribution to $m'$ that is coming from the $(\mathbb Z/q\mathbb Z)^{\times}$ factor will lead to additional powers of $q$ in the discriminant.
As a more precise example, suppose that $e = 1$, and write $(\mathbb Z/q\mathbb Z)^{\times} = H \times H',$ where $H$ has order prime-to-$p$ and $H'$ has order $p$. Consider two possibilities for $Gal(K/Q)$. In the first case, suppose that $Gal(K/Q) = (\mathbb Z/p^2\mathbb Z)^{\times} \times H$, while in the second case suppose that $Gal(K/Q) = (\mathbb Z/p\mathbb Z)^{\times}\times (\mathbb Z/q\mathbb Z)^{\times}$. In each case $Gal(K/Q)$ is the product of a cyclic group of order $(p-1)$ and a cyclic group of order $(q-1)$.
But in the first case the discriminant (up to sign) will be $p^{2p^2 - 3p} q^{(q-p-1)/p}$, while in the second case it will be $p^{p-2}q^{q-2}$.
And in the first case we have that $K \subset \mathbb Q(\zeta_{p^2},\zeta_q)$ (but no smaller cyclotomic field), while in the second case $K = \mathbb Q(\zeta_p,\zeta_q)$.
I didn't try to work this method out more systematically, but presumably one can, and certainly it shouldn't be so bad to apply by hand to any particular small example that you have in mind.