Here is a partial affirmative answer using mod 2 Steenrod operations; the simplest case of this (for $n$ and $k$ even) is just a correction of the slightly incorrect answer originally posted by Włodzimierz Holsztyński. The result is that if $k+1$ and $n+1$ are both odd multiples of $2^d$ for some integer $d\geq 0$, then $\mathbb{C}P^n/\mathbb{C}P^k$ has the fixed point property. In particular, for $d=0$ we get the fixed point property whenever $n$ and $k$ are both even. All cohomology in this answer will have coefficients in $\mathbb{F}_2$.

Let's start by describing the action of the Steenrod squares on the cohomology $H^*(\mathbb{C}P^n)=\mathbb{F}_2[x]/(x^{n+1})$. The following formulas are easy to prove by induction using the Cartan formula (induct on $d$ and for fixed $d$ induct on $m$):

$$Sq^{2^{d+1}}\left(x^{2^dm}\right)=x^{2^d(m+1)} \text{ if $m$ is odd}$$
$$Sq^{2^{d+1}}\left(x^{2^dm}\right)=0 \text{ if $m$ is even}$$

From these, we deduce the following for all $0\leq \ell<2^d$:
$$Sq^{2^{d+1}}\left(x^{2^dm+\ell}\right)=x^{2^d(m+1)+\ell} \text{ if $m$ is odd}$$
$$Sq^{2^{d+1}}\left(x^{2^dm+\ell}\right)=0 \text{ if $m$ is even}$$

The quotient map $\mathbb{C}P^n\to\mathbb{C}P^n/\mathbb{C}P^k$ identifies $H^*(\mathbb{C}P^n/\mathbb{C}P^k)$ with the subring of $H^*(\mathbb{C}P^n)$ which has as a basis $\{1,x^{k+1},x^{k+2},\dots, x^n\}$, and so the same relations hold in $H^*(\mathbb{C}P^n/\mathbb{C}P^k)$.

Suppose now that $f:\mathbb{C}P^n/\mathbb{C}P^k\to\mathbb{C}P^n/\mathbb{C}P^k$ is any map. For $k<i\leq n$, let $a_i\in \mathbb{F}_2$ be such that $f^*(x^i)=a_ix^i$. By the Lefschetz fixed point theorem, $f$ must have a fixed point if $1+\sum_{k+1}^n a_i\neq 0$ (the $1$ coming from $H^0$), or equivalently if $\sum a_i=0$.

Since $f^*$ must commute with Steenrod operations, we must have $a_{2^dm+\ell}=a_{2^d(m+1)+\ell}$ for $m$ odd and $0\leq \ell<2^d$, as long as $k<2^dm+\ell<2^d(m+1)+\ell\leq n$. Together, these relations imply that if $m$ is odd and $k<2^dm<2^d(m+2)-1\leq n$, then all the $a_i$ for $2^dm\leq i \leq 2^d(m+2)-1$ are equal to each other (everything below $2^d(m+1)$ can be related to $2^d(m+1)$ using the smaller Steenrod squares, and everything above $2^d(m+1)$ can be related to something below it using $Sq^{2^{d+1}}$). That is, the $a_i$ are constant in blocks of length $2^{d+1}$ starting from an odd multiple of $2^d$.

Now suppose that $k+1$ and $n+1$ are both odd multiples of $2^d$. The numbers from $k+1$ to $n$ can be broken into blocks of length $2^{d+1}$, each starting with an odd multiple of $2^d$. All of the $a_i$ in each block are equal to each other, and hence their sum is zero since there are an even number of them. Thus the sum of all of the $a_i$ is zero, and so $f$ must have a fixed point.

Let me conclude with a couple remarks on this result. First, as Włodzimierz observed, this argument works equally well for projective spaces over $\mathbb{R}$ or $\mathbb{H}$ (for $\mathbb{R}$, replace $Sq^{2^{d+1}}$ with $Sq^{2^d}$ and for $\mathbb{H}$ replace it with $Sq^{2^{d+2}}$). Second, the condition obtained here is sufficient but not necessary for $\mathbb{C}P^n/\mathbb{C}P^k$ to have the fixed point property. Indeed, in the comments I sketched an argument using cup products and integer coefficients rather than Steenrod squares and mod 2 coefficients which shows that the fixed point property holds when $n\gg k$ as long as either $n$ is even or $k$ is odd (note that in fact, using only mod 2 coefficients there is no hope of proving the fixed point property in cases where $n$ and $k$ have different parity).