In the sequel we will consider $P$ being made of the primes $1 < p_1 < p_2 < \dots < p_k$, with $k = |P| \ge 1$.

a. By the Chinese Remainder Theorem there will exist some $a \in \mathbb{N}$ such that $a \equiv -i \pmod{p_i}$, hence $p_i \mid a + i$. Then the set $\{a + i; i = 1, 2, \dots, k\}$ of $k$ consecutive integers has the desired property, hence $m(P) \ge k$. When $\min P > |P|$, within any set of $|P| + 1$ consecutive integers at most one is divisible by any $p \in P$, hence by the Pigeonhole Principle there will be one not divisible by any of the primes in $P$. On the other hand, when $\min P \le |P|$, we will make again use of the Chinese Remainder Theorem, so there will exist some $a \in \mathbb{N}$ such that $a \equiv -r_i \pmod{p_i}$, hence $p_i \mid a+r_i$, where $\{r_i; i = 1, 2, \dots, k\} = \{1, 2, \dots, k\}$ and the extra requirement that $r_1 = k+1-p_1$. It follows that the set $\{a+i; i = 1, 2, \dots, k, k+1\}$ of $k+1$ consecutive integers has the desired property, hence $m(P) \ge k+1 > |P|$.

b. Now, let a set made of $m$ consecutive integers have the desired property. For any $\emptyset \ne I \subseteq \{1, 2, \dots, k\}$, the number $N(I)$ of its elements which are divisible by $\prod_{i \in I} p_i$ will satisfy the inequality
$$\frac{m}{\prod_{i \in I} p_i} - 1 < \left\lfloor \frac{m}{\prod_{i \in I} p_i} \right\rfloor \le N(I) \le \left\lceil \frac{m}{\prod_{i \in I} p_i} \right\rceil < \frac{m}{\prod_{i \in I} p_i} + 1.$$
Then, by the Principle of Inclusion/Exclusion, one has
$$m = \sum_{i=1}^{k} (-1)^{i+1} \sum_{|I|=i} N(I) < \sum_{i=1}^{k} \binom{k}{i} + m \sum_{i=1}^{k} (-1)^{i+1} \sum_{|I|=i} \frac{1}{\prod_{i \in I} p_i}.$$
The first term is clearly equal to $2^k - 1$, while the second is equal to
$$m \left( 1 - \prod_{i=1}^{k} \left( 1 - \frac{1}{p_i} \right) \right) \le m \left( 1 - \prod_{i=1}^{k} \left( 1 - \frac{1}{i+1} \right) \right) = m - \frac{m}{k+1},$$
therefore $m < (k+1)(2^k - 1)$, and so will be $m(P)$.

Alternative Solution (ii). (F. Chindea)

Not only an elegant alternative approach, but vastly improving on the required bound. Start by proving the following

LEMMA. If the integer arithmetic sequences $\{mn_j\}_{m \in \mathbb{Z}}$, $1 \le j \le k$, are such that they cover $2^k$ consecutive integers, then they cover $\mathbb{Z}$.
(Particular case of a Crittenden & Vanden Eynden Theorem)

Proof. Let $\omega_j = e^{n_j}$ be a primitive root of unity. Then $n_j \mid m$ if and only if $\omega_j^m = 1$, hence $m$ is divisible by at least one $n_j$ if and only if $0 = \prod_{j=1}^k (1 - \omega_j^m) = \sum_{\emptyset \subseteq S \subseteq [k]} (-1)^{|S|} e^{2m\pi i \sum_{j \in S} \frac{1}{n_j}}$. Denote $\alpha_S = (-1)^{|S|}$, $z_S = e^{2\pi i \sum_{j \in S} \frac{1}{n_j}}$, $u_m = \sum_{\emptyset \subseteq S \subseteq [k]} \alpha_S z_S^m$.

Consider the polynomial $f(x) = \prod_{\emptyset \subseteq S \subseteq [k]} (x - z_S) = \sum_{j=0}^{2^k} c_j x^j$, of degree $\deg f = 2^k$, and with $c_{2^k} = 1$, $|c_0| = \prod_{\emptyset \subseteq S \subseteq [k]} |z_S| = 1$. Then $\alpha_S z_S^b f(z_S) = 0$ for any integer $b$, hence
$$0 = \sum_{\emptyset \subseteq S \subseteq [k]} \alpha_S z_S^b f(z_S) = \sum_{j=0}^{2^k} c_j \sum_{\emptyset \subseteq S \subseteq [k]} \alpha_S z_S^{b+j} = \sum_{j=0}^{2^k} c_j u_{b+j}.$$
Assume the $2^k$ consecutive integers $a+1, \dots, a+2^k$ are covered, so, according with the above, $u_{a+1} = \dots = u_{a+2^k} = 0$. Since $c_0 \neq 0$ and $c_{2^k} \neq 0$, by simple induction one gets $u_b = 0$ for any integer $b$ (a linear recurrence relation of order $2^k$ containing $2^k$ consecutive null terms generates an all-null sequence). $\square$

Now, in our case, if we assume $m(P) \ge 2^{|P|}$, then the $|P|$ arithmetic sequences given by $\{mp\}_{m \in \mathbb{Z}}$, $p \in P$, will cover $2^k$ consecutive integers, hence they will cover $\mathbb{Z}$. But clearly a large enough prime $q$ is not divisible by any $p \in P$, so this is absurd, therefore $m(P) < 2^{|P|}$.