1. Let the counts of digits be:

   • $x$ = number of 1's
   • $y$ = number of 2's
   • $z$ = number of 3's

   Since the number is a 7-digit number formed only from digits $1,2,3$: $x+y+z=7$

   Also, the sum of digits is 11: $x+2y+3z=11$

2. Eliminate $x$ using $x=7-y-z$: $7-y-z+2y+3z=11$ $7+y+2z=11$ $y+2z=4$

3. Find all non-negative integer solutions of $y+2z=4$

   Possible values:

   • If $z=0$, then $y=4$, so $x=7-4-0=3$
   • If $z=1$, then $y=2$, so $x=7-2-1=4$
   • If $z=2$, then $y=0$, so $x=7-0-2=5$

   So the possible distributions are:

   • $(x,y,z)=(3,4,0)$
   • $(x,y,z)=(4,2,1)$
   • $(x,y,z)=(5,0,2)$

4. Count numbers for each case using permutations of repeated digits.

   Case 1: $(3,4,0)$

   Digits: three 1's and four 2's $\frac{7!}{3!4!}=35$

   Case 2: $(4,2,1)$

   Digits: four 1's, two 2's, one 3 $\frac{7!}{4!2!1!}=\frac{5040}{48}=105$

   Case 3: $(5,0,2)$

   Digits: five 1's and two 3's $\frac{7!}{5!2!}=21$

5. Total number of such 7-digit numbers: $35+105+21=161$

6. Compare with given options: Option C is $161$.

Therefore, the number of elements in $P$ is: $\boxed{161}$