1. We need to maximize $P=a^5b^3c^2d$ subject to $a+b+c+d=11,$ where $a,b,c,d>0$.

2. Use the weighted AM-GM inequality.

Notice that the total exponent sum is $5+3+2+1=11.$ So write $\frac{5a+3b+2c+d}{11} \ge \sqrt[11]{a^5b^3c^2d}.$

Since $5a+3b+2c+d \neq a+b+c+d,$ this form is not directly useful. Instead, for a fixed sum $a+b+c+d=11$, the standard result for maximizing $a^{\alpha}b^{\beta}c^{\gamma}d^{\delta}$ with positive variables and fixed sum says the maximum occurs when $a:b:c:d=\alpha:\beta:\gamma:\delta=5:3:2:1.$

3. Let $a=5k,\quad b=3k,\quad c=2k,\quad d=k.$ Then $a+b+c+d=5k+3k+2k+k=11k=11,$ so $k=1.$ Hence, $a=5,\quad b=3,\quad c=2,\quad d=1.$

4. Now compute the maximum value: $P_{\max}=5^5\cdot 3^3\cdot 2^2\cdot 1.$ Calculate step by step: $5^5=3125,\quad 3^3=27,\quad 2^2=4.$ Thus $P_{\max}=3125\cdot 27\cdot 4.$ First, $27\cdot 4=108,$ so $P_{\max}=3125\cdot 108.$ Since the question says this maximum value is $3750\beta,$ we get $3750\beta=3125\cdot 108.$ Therefore, $\beta=\frac{3125\cdot 108}{3750}.$ Now, $\frac{3125}{3750}=\frac{5}{6},$ so $\beta=108\cdot \frac{5}{6}=90.$

5. Therefore, $\boxed{\beta=90}.$ So the correct option is C.