There are $2^M\binom{N}{M}$ ways of pulling $M$ socks without a pair, so the probability for that is
$$p(N,M)=2^M\frac{\binom NM}{\binom{2N}M}\;,$$
where the relationship between $p$ and your $P$ is $1-p=P\;\%$. This can't be solved for $M$ in closed form for arbitrary $p$ and $N$, but asymptotically for $N$ large and $M\ll N$ we have
$$
\begin{eqnarray}
p(N,M)
&=&
2^M\frac{\binom NM}{\binom{2N}M}
\\
&=&
2^M\frac{N!(2N-M)!}{(2N)!(N-M)!}
\\
&\approx&
2^M\frac{N^N(2N-M)^{2N-M}}{(2N)^{2N}(N-M)^{N-M}}\sqrt{\frac{N(2N-M)}{2N(N-M)}}
\\
&=&
\frac{\left(1-\frac M{2N}\right)^{2N-M+1/2}}{\left(1-\frac MN\right)^{N-M+1/2}}
\\
&=&
\frac{\left(1-\frac M{N}+\left(\frac{M}{2N}\right)^2\right)^{N-M/2+1/4}}{\left(1-\frac MN\right)^{N-M+1/2}}
\\
&=&
\left(1+\left(\frac{M}{2N}\right)^2/\left(1-\frac MN\right)\right)^{N-M/2+1/4}\left(1-\frac MN\right)^{M/2-1/4}
\\
&\approx&
\left(1+\frac{M^2}{4N}\right)\left(1-\left(\frac M2-\frac14\right)\frac MN\right)
\\
&\approx&
1-\frac{M(M-1)}{4N}\;.
\end{eqnarray}
$$
So the probability of pulling a sock with a pair is about $M(M-1)/(4N)$. This approximation is good enough for some practical cases; for instance, if you pull $4$ out of $20$ socks, the exact probability of finding a pair is $15.\overline{176715}\%$ and the approximation yields $15\%$.
However, this approximation breaks down if $M$ is of the order of $\sqrt N$, which will be the case if you want a probability near $50\%$. For this case, we need a slightly different approximation. Generalizing $p$ to fractional $M$ by replacing $k!$ by $\Gamma(k+1)$, we have
$$
\begin{eqnarray}
\lim_{N\to\infty}
p(N,\sqrt{\lambda N})
&=&
\lim_{N\to\infty}2^{\sqrt{\lambda N}}\frac{\Gamma(N+1)\Gamma(2N-\sqrt{\lambda N}+1)}{\Gamma(2N+1)\Gamma(N-\sqrt{\lambda N}+1)}
\\
&=&
\lim_{N\to\infty}
\left(1+\left(\frac{\sqrt{\lambda N}}{2N}\right)^2/\left(1-\frac{\sqrt{\lambda N}}N\right)\right)^{N-\sqrt{\lambda N}/2+1/4}\left(1-\frac{\sqrt{\lambda N}}N\right)^{\sqrt{\lambda N}/2-1/4}
\\
&=&
\lim_{N\to\infty}
\left(1+\left(\frac{\sqrt{\lambda N}}{2N}\right)^2\right)^{N}\left(1-\frac{\sqrt{\lambda N}}N\right)^{\sqrt{\lambda N}/2}
\\
&=&
\lim_{N\to\infty}
\left(1+\frac{\lambda}{4N}\right)^{N}\left(1-\frac{\lambda/2}{\sqrt{\lambda N}/2}\right)^{\sqrt{\lambda N}/2}
\\
&=&
\mathrm e^{\lambda/4}\mathrm e^{-\lambda/2}
\\
&=&
\mathrm e^{-\lambda/4}\;.
\end{eqnarray}
$$
So for large $N$, we need to pull $\sqrt{\lambda N}$ socks to get a probability of $\mathrm e^{-\lambda/4}$ not to get a pair. This we can readily solve for $\lambda$ for given limit probability $p$ of not getting a pair:
$$p=\mathrm e^{-\lambda/4} \rightarrow \lambda=-4\log p\;.$$
For instance, for a $50\%$ chance, we get $\lambda =4\log2$, so $\sqrt{\lambda}=2\sqrt{\log2}\approx1.665$. That is, for large $N$, to have a $50\%$ chance of pulling a pair, you need to pull about $1.665\sqrt N$ socks. Since $1/p$ is the average number $n$ of times you need to pull to not get a pair, we can also write this as
$$M=\sqrt{\lambda N}=\sqrt{-4N\log p}=\sqrt{4N\log1/p}=2\sqrt{N\log n}\;.$$
So $M$ is twice the geometric mean of the number of socks and the logarithm of the average number of times you want to be able to do this until the first miss.
