We have $\binom n2$ pairs of people, for each pair the probability to share the birthday being $d^{-1}$. Hence, denoting by $K$ the number of birthday-sharing pairs, the expectation of $K$ is $d^{-1}\binom n2$, and by Markov's inequality, the probability to have $K\ge k$ is less than $n^2/(2kd)$. Thus, in order to have at least $k$ pairs with probability $0.5$, one needs $n>\sqrt{kd}$.
On the other hand, denoting for each $j\in[1,d]$ by $n_j$ the number of people having their birthday at the $j$th day of the year, by Jensen's inequality we have
$$ K = \binom{n_1}2+\dotsb+\binom{n_d}2 \ge d\binom{n/d}2; $$
therefore, if $n\ge\sqrt{2kd}+d$, then $K\ge n(n/d-1)/2>k$, meaning that in this case we have at least $k$ pairs of people with the same birthday with probability $1$.
To summarize,
The smallest number of people needed to ensure at least $k$ birthday-sharing pairs with probability $0.5$ satisfies
$$ \sqrt{kd} < n < \sqrt{2kd}+d. $$
In the regime where $d=o(k^{1/3})$, a precise asymptotic can be given showing that the number of people needed is
$$ n = (1+o(1))\sqrt{2kd}. $$
To see this, observe that each $n_j$ (introduced above) is distributed as $B(n,d^{-1})$. Hence, by the Hoeffding's inequality, for any $\epsilon>0$, we have $n(d^{-1}-\epsilon)<n_j<n(d^{-1}+\epsilon)$ with probability a least $1-2e^{-2\epsilon^2n}$. By the union bound, with probability at least $1-2de^{-2\epsilon^2n}$, we then have $n(d^{-1}-\epsilon)<n_j<n(d^{-1}+\epsilon)$ for each $j\in[1,d]$, resulting in $d\binom{n(d^{-1}-\epsilon)}{2}\le K\le d\binom{n(d^{-1}+\epsilon)}{2}$. The assertion now follows by letting, say, $\epsilon:=d^{-1/2}k^{-1/6}$ and observing that for $n=(1+o(1))\sqrt{2kd}$ we have then $e^{-\epsilon^2n}=o(1)$ and $\epsilon=o(d^{-1})$.
