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Edit:Let $K=\mathbb Q(\zeta_p)^+$

Since it's easy to show $-1 \in N(\mathcal O_K ^{\times})$ and $\mathrm{Gal}(K/k)$ is cyclic, your question is equivalent to ask whether $\mathrm{H}^2(K/k,\mathcal O_K^\times)=0$. By a fact about Herbrand quotient ([1]Proposition1.2.4) , this is equivalent to ask whether $\mathrm{H}^1(K/k,\mathcal O_K^\times)=n$, where $n=[K:k]$. Apply the exact sequence([1]Proposition1.2.3) to our case $K/k$. Since $K/k$ is totally ramified, and the raimified primes are principal. We have the following exact sequence $$0\longrightarrow \mathrm{Ker}(J) \longrightarrow \mathrm{H}^1(K/k,\mathcal O_K^\times) \longrightarrow \mathbb{Z}/n\mathbb{Z}\longrightarrow 0,$$

where $J$ is the natural map from $Cl(k)$ to $Cl(K)$. Then we know $|\mathrm{H}^2(K/k,\mathcal O_K^\times)|=|\mathrm{Ker}(J)|$. So your question is to ask whether there is a nonprincipal ideal of $k$ becomes principal in $K$. If $h_k$ is coprime to $n$, then $J$ is injective, so the norm of units is surjective as Pound Sterling said. If $h_k$ is not coprime to $n$, $J$ may be injective or not. For example, $p=229$,$|\mathrm{Ker}J|=3$, as Pound Sterling says the norm index is $3$. $p=2089,h_k=3$, and $\mathrm{gcd}(h_k,n)=3$, but $J$ is injective, see[2,Page 2728], so the norm map between units is surjective. Numerically, most $p$ such that $\mathrm{gcd}(h_k,n)>1$ adimt a nontrival $\mathrm{Ker} J$, hence the norm map between units is not surjective. See the discussion in [2]Page 2727.

References:[1] Topics in Iwasawa theory. Greenberg.

[2] Visibility of ideal classes. Schoof and Washington.

Edit:Let $K=\mathbb Q(\zeta_p)^+$

Since it's easy to show $-1 \in N(\mathcal O_K ^{\times})$ and $\mathrm{Gal}(K/k)$ is cyclic, your question is equivalent to ask whether $\mathrm{H}^2(K/k,\mathcal O_K^\times)=0$. By a fact about Herbrand quotient ([1]Proposition1.2.4) , this is equivalent to ask whether $|\mathrm{H}^1(K/k,\mathcal O_K^\times)|=n$, where $n=[K:k]$. Apply the exact sequence([1]Proposition1.2.3) to our case $K/k$. Since $K/k$ is totally ramified, and the raimified primes are principal. We have the following exact sequence $$0\longrightarrow \mathrm{Ker}(J) \longrightarrow \mathrm{H}^1(K/k,\mathcal O_K^\times) \longrightarrow \mathbb{Z}/n\mathbb{Z}\longrightarrow 0,$$

where $J$ is the natural map from $Cl(k)$ to $Cl(K)$. Then we know $|\mathrm{H}^2(K/k,\mathcal O_K^\times)|=|\mathrm{Ker}(J)|$. So your question is to ask whether there is a nonprincipal ideal of $k$ becomes principal in $K$. If $h_k$ is coprime to $n$, then $J$ is injective, so the norm of units is surjective as Pound Sterling said. If $h_k$ is not coprime to $n$, $J$ may be injective or not. For example, $p=229$,$|\mathrm{Ker}J|=3$, as Pound Sterling says the norm index is $3$. $p=2089,h_k=3$, and $\mathrm{gcd}(h_k,n)=3$, but $J$ is injective, see[2,Page 2728], so the norm map between units is surjective. Numerically, most $p$ such that $\mathrm{gcd}(h_k,n)>1$ adimt a nontrival $\mathrm{Ker} J$, hence the norm map between units is not surjective. See the discussion in [2]Page 2727.

References:[1] Topics in Iwasawa theory. Greenberg.

[2] Visibility of ideal classes. Schoof and Washington.

Edit:Let $K=\mathbb Q(\zeta_p)^+$

Since it's easy to show $-1 \in N(\mathcal O_K ^{\times})$ and $\mathrm{Gal}(K/k)$ is cyclic, your question is equivalent to ask whether $\mathrm{H}^2(K/k,\mathcal O_K^\times)=0$. By a well-known fact about Herbrand quotient ([1]Proposition1.2.4) , this is equivalent to ask whether $\mathrm{H}^1(K/k,\mathcal O_K^\times)=n$, where $n=[K:k]$. Apply an easy exact sequence([1]Proposition1.2.3) to our case $K/k$. Since $K/k$ is totally ramified, and the raimified primes are principal. It's easy to show that we have the following exact sequence $$0\longrightarrow \mathrm{Ker}(J) \longrightarrow \mathrm{H}^1(K/k,\mathcal O_K^\times) \longrightarrow \mathbb{Z}/n\mathbb{Z}\longrightarrow 0,$$

where $J$ is the natural map from $Cl(k)$ to $Cl(K)$. Then we know $|\mathrm{H}^2(K/k,\mathcal O_K^\times)|=|\mathrm{Ker}(J)|$. So your question is to ask whether there is a nonprincipal ideal of $k$ becomes principal in $K$. It's trivial to show that if $h_k$ is coprime to $n$, then $J$ is injective, so the norm of units is surjective as Pound Sterling says. If $h_k$ is not coprime to $n$, $J$ may be injective or not. For example, $p=229$,$|\mathrm{Ker}J|=3$, as Pound Sterling says the norm index is $3$. $p=2089,h_k=3$, and $\mathrm{gcd}(h_k,n)=3$, but $J$ is injective, see[2,Page 2728], so the norm map between units is surjective. Numerically, most $p$ such that $\mathrm{gcd}(h_k,n)>1$ adimt a nontrival $\mathrm{Ker} J$, hence the norm map between units is not surjective. See the discussion in [2]Page 2727.

References:[1] Topics in Iwasawa theory. Greenberg.

[2] Visibility of ideal classes. Schoof and Washington.

Edit:Let $K=\mathbb Q(\zeta_p)^+$

Since it's easy to show $-1 \in N(\mathcal O_K ^{\times})$ and $\mathrm{Gal}(K/k)$ is cyclic, your question is equivalent to ask whether $\mathrm{H}^2(K/k,\mathcal O_K^\times)=0$. By a fact about Herbrand quotient ([1]Proposition1.2.4) , this is equivalent to ask whether $\mathrm{H}^1(K/k,\mathcal O_K^\times)=n$, where $n=[K:k]$. Apply the exact sequence([1]Proposition1.2.3) to our case $K/k$. Since $K/k$ is totally ramified, and the raimified primes are principal. We have the following exact sequence $$0\longrightarrow \mathrm{Ker}(J) \longrightarrow \mathrm{H}^1(K/k,\mathcal O_K^\times) \longrightarrow \mathbb{Z}/n\mathbb{Z}\longrightarrow 0,$$

where $J$ is the natural map from $Cl(k)$ to $Cl(K)$. Then we know $|\mathrm{H}^2(K/k,\mathcal O_K^\times)|=|\mathrm{Ker}(J)|$. So your question is to ask whether there is a nonprincipal ideal of $k$ becomes principal in $K$. If $h_k$ is coprime to $n$, then $J$ is injective, so the norm of units is surjective as Pound Sterling said. If $h_k$ is not coprime to $n$, $J$ may be injective or not. For example, $p=229$,$|\mathrm{Ker}J|=3$, as Pound Sterling says the norm index is $3$. $p=2089,h_k=3$, and $\mathrm{gcd}(h_k,n)=3$, but $J$ is injective, see[2,Page 2728], so the norm map between units is surjective. Numerically, most $p$ such that $\mathrm{gcd}(h_k,n)>1$ adimt a nontrival $\mathrm{Ker} J$, hence the norm map between units is not surjective. See the discussion in [2]Page 2727.

References:[1] Topics in Iwasawa theory. Greenberg.

[2] Visibility of ideal classes. Schoof and Washington.

Since it's easy to show $-1 \in N(\mathcal O_K ^{\times})$ and $\mathrm{Gal}(K/k)$ is cyclic, your question is equivalent to ask whether $\mathrm{H}^2(K/k,\mathcal O_K^\times)=0$. By a well-known fact about Herbrand quotient ([1]Proposition1.2.4) , this is equivalent to ask whether $\mathrm{H}^1(K/k,\mathcal O_K^\times)=n$, where $n=[K:k]$. Apply an easy exact sequence([1]Proposition1.2.3) to our case $K/k$. Since $K/k$ is totally ramified, and the raimified primes are principal. It's easy to show that we have the following exact sequence $$0\longrightarrow \mathrm{Ker}(J) \longrightarrow \mathrm{H}^1(K/k,\mathcal O_K^\times) \longrightarrow \mathbb{Z}/n\mathbb{Z}\longrightarrow 0,$$

where $J$ is the natural map from $Cl(k)$ to $Cl(K)$. Then we know $|\mathrm{H}^2(K/k,\mathcal O_K^\times)|=|\mathrm{Ker}(J)|$. So your question is to ask whether there is a nonprincipal ideal of $k$ becomes principal in $K$. It's trivial to show that if $h_k$ is coprime to $n$, then $J$ is injective, so the norm of units is surjective as Pound Sterling says. If $h_k$ is not coprime to $n$, $J$ may be injective or not. For example, $p=229$,$|\mathrm{Ker}J|=3$, as Pound Sterling says the norm index is $3$. $p=2089,h_k=3$, and $\mathrm{gcd}(h_k,n)=3$, but $J$ is injective, see[2,Page 2728], so the norm map between units is surjective. Numerically, most $p$ such that $\mathrm{gcd}(h_k,n)>1$ adimt a nontrival $\mathrm{Ker} J$, hence the norm map between units is not surjective. See the discussion in [2]Page 2727.

References:[1] Topics in Iwasawa theory. Greenberg.

[2] Visibility of ideal classes. Schoof and Washington.

Edit:The $K$ in my answer is $\mathbb Q(\zeta_p)^+$

Edit:Let $K=\mathbb Q(\zeta_p)^+$

Since it's easy to show $-1 \in N(\mathcal O_K ^{\times})$ and $\mathrm{Gal}(K/k)$ is cyclic, your question is equivalent to ask whether $\mathrm{H}^2(K/k,\mathcal O_K^\times)=0$. By a well-known fact about Herbrand quotient ([1]Proposition1.2.4) , this is equivalent to ask whether $\mathrm{H}^1(K/k,\mathcal O_K^\times)=n$, where $n=[K:k]$. Apply an easy exact sequence([1]Proposition1.2.3) to our case $K/k$. Since $K/k$ is totally ramified, and the raimified primes are principal. It's easy to show that we have the following exact sequence $$0\longrightarrow \mathrm{Ker}(J) \longrightarrow \mathrm{H}^1(K/k,\mathcal O_K^\times) \longrightarrow \mathbb{Z}/n\mathbb{Z}\longrightarrow 0,$$

where $J$ is the natural map from $Cl(k)$ to $Cl(K)$. Then we know $|\mathrm{H}^2(K/k,\mathcal O_K^\times)|=|\mathrm{Ker}(J)|$. So your question is to ask whether there is a nonprincipal ideal of $k$ becomes principal in $K$. It's trivial to show that if $h_k$ is coprime to $n$, then $J$ is injective, so the norm of units is surjective as Pound Sterling says. If $h_k$ is not coprime to $n$, $J$ may be injective or not. For example, $p=229$,$|\mathrm{Ker}J|=3$, as Pound Sterling says the norm index is $3$. $p=2089,h_k=3$, and $\mathrm{gcd}(h_k,n)=3$, but $J$ is injective, see[2,Page 2728], so the norm map between units is surjective. Numerically, most $p$ such that $\mathrm{gcd}(h_k,n)>1$ adimt a nontrival $\mathrm{Ker} J$, hence the norm map between units is not surjective. See the discussion in [2]Page 2727.

References:[1] Topics in Iwasawa theory. Greenberg.

[2] Visibility of ideal classes. Schoof and Washington.