No, $\mathbb C\mathbb P^n \setminus K$ is never a retract of \mathbb C \mathbb P^{n+1} \setminus K$.

If $K$ is nonempty, the generator $x$ of $H^2( \mathbb C \mathbb P^n\setminus K )$ satisfies $x^n=0$. So its pullback along any map $\mathbb C \mathbb P^{n+1} \setminus K \to \mathbb C \mathbb P^n \setminus K$ must be a class $y$ satisfying $y^n=0$.

Since $H^i(\mathbb C \mathbb P^{n+1} \setminus K ) \to H^i(\mathbb C \mathbb P^{n+1})$ is an isomorphism for $i=2, 2n$, any such class must be $0$: It must be a pullback from $\mathbb C \mathbb P^{n+1}$, and if it is nonzero then it is the pullback of a nonzero class so the $n$th power of the pullback is nonzero, thus the pullback of the $n$th power is nonzero, hence the $n$th power is nonzero.

But if $X$ is a retract of $Y$, then $X \to Y \to X$ is an isomorphism, so $H^i (X) \to H^i(Y)$ must be injective, so the pullback of $x$ cannot be zero. This is a contradiction, so it's not a retract.

However, $\mathbb C \mathbb P^n$ is a retract of $\mathbb C \mathbb P^{n+1} \setminus K$ for any nonempty $K$, by embedding as a hyperplane disjoint from $K$ and then projecting from a point in $K$.

No, $\mathbb{CP}^n \setminus K$ is never a retract of $\mathbb{CP}^{n+1} \setminus K$.

If $K$ is nonempty, the generator $x$ of $H^2( \mathbb {CP}^n\setminus K )$ satisfies $x^n=0$. So its pullback along any map $\mathbb{CP}^{n+1} \setminus K \to \mathbb{CP}^n \setminus K$ must be a class $y$ satisfying $y^n=0$.

Since $H^i(\mathbb{CP}^{n+1} \setminus K ) \to H^i(\mathbb {CP}^{n+1})$ is an isomorphism for $i=2, 2n$, any such class must be $0$: It must be a pullback from $\mathbb {CP}^{n+1}$, and if it is nonzero then it is the pullback of a nonzero class so the $n$th power of the pullback is nonzero, thus the pullback of the $n$th power is nonzero, hence the $n$th power is nonzero.

But if $X$ is a retract of $Y$, then $X \to Y \to X$ is an isomorphism, so $H^i (X) \to H^i(Y)$ must be injective, so the pullback of $x$ cannot be zero. This is a contradiction, so it's not a retract.

However, $\mathbb {CP}^n$ is a retract of $\mathbb {CP}^{n+1} \setminus K$ for any nonempty $K$, by embedding as a hyperplane disjoint from $K$ and then projecting from a point in $K$.

If $K$ is nonempty, the generator $x$ of $H^2( \mathbb C \mathbb P^n\setminus K )$ satisfies $x^n=0$. So its pullback along any map $\mathbb C \mathbb P^{n+1} \setminus K \to \mathbb C \mathbb P^n \setminus K$ must be a class $y$ satisfying $y^n=0$.

Since $H^i(\mathbb C \mathbb P^{n+1} \setminus K ) \to H^i(\mathbb C \mathbb P^{n+1})$ is an isomorphism for $i=2, 2n$, any such class must be $0$: It must be a pullback from $\mathbb C \mathbb P^{n+1}$, and if it is nonzero then it is the pullback of a nonzero class so the $n$th power of the pullback is nonzero, thus the pullback of the $n$th power is nonzero, hence the $n$th power is nonzero.

But if $X$ is a retract of $Y$, then $X \to Y \to X$ is an isomorphism, so $H^i (X) \to H^i(Y)$ must be injective, so the pullback of $x$ cannot be zero. This is a contradiction, so it's not a retract.

However, $\mathbb C \mathbb P^n$ is a retract of $\mathbb C \mathbb P^{n+1} \setminus K$ for any nonempty $K$, by embedding as a hyperplane disjoint from $K$ and then projecting from a point in $K$.

No, $\mathbb C\mathbb P^n \setminus K$ is never a retract of \mathbb C \mathbb P^{n+1} \setminus K$.

If $K$ is nonempty, the generator $x$ of $H^2( \mathbb C \mathbb P^n\setminus K )$ satisfies $x^n=0$. So its pullback along any map $\mathbb C \mathbb P^{n+1} \setminus K \to \mathbb C \mathbb P^n \setminus K$ must be a class $y$ satisfying $y^n=0$.

Since $H^i(\mathbb C \mathbb P^{n+1} \setminus K ) \to H^i(\mathbb C \mathbb P^{n+1})$ is an isomorphism for $i=2, 2n$, any such class must be $0$: It must be a pullback from $\mathbb C \mathbb P^{n+1}$, and if it is nonzero then it is the pullback of a nonzero class so the $n$th power of the pullback is nonzero, thus the pullback of the $n$th power is nonzero, hence the $n$th power is nonzero.

But if $X$ is a retract of $Y$, then $X \to Y \to X$ is an isomorphism, so $H^i (X) \to H^i(Y)$ must be injective, so the pullback of $x$ cannot be zero. This is a contradiction, so it's not a retract.

However, $\mathbb C \mathbb P^n$ is a retract of $\mathbb C \mathbb P^{n+1} \setminus K$ for any nonempty $K$, by embedding as a hyperplane disjoint from $K$ and then projecting from a point in $K$.