Monty hall problem, getting different probabilities using different formulas? In my Monty hall problem, I am computing what is the probability that P(H=1|D=3) i.e. price is behind door 1 and the 3rd door is opened.

$P(H=1|D=3) = p(H=1) * \frac{p(D=3|H=1)} {p(D=3)} = 1/3 * 1/2 / 1/3 = 1/2 = 50\%$
$P(H=1|D=3) = p(H=1)  \frac{p(D=3|H=1}{\sum_{i=1} ^3 p(H=i) * p(D=3|H=i)} = \frac{1/3 * 1/2} { 1/3 * 1/2 + 1/3 * 1 + 0} = 1/3 = 33\% $

And when I use Bayes formula without summation in the denominator I get answer 50% and when I use summation in the denominator in Bayes formula I get 33%. Why there is difference? 

Your problem is that $P(D=3 \mid H=2) = \frac{1}{2}$ but you incorrectly wrote that it equals 1
Explanation:
Let $D$ be the door with the prize. Let $H$ be the door Mr. Hall opens. The joint distribution of $D$ and $H$ is described in this table:
$$ \begin{array}{cccc} &\text{H=1}&\text{H=2}&\text{H=3} \\
D = 1 & 0 & \frac{1}{6} & \frac{1}{6} \\
D = 2 & \frac{1}{6} & 0  & \frac{1}{6} \\
D = 3 & \frac{1}{6} & \frac{1}{6} & 0 
\end{array}$$
The problem is in the denominator of your second formula:

$P(H=1|D=3) = p(H=1)  \frac{p(D=3|H=1}{\sum_{i=1} ^3 p(H=i) * p(D=3|H=i)} = \frac{1/3 * 1/2} { 1/3 * 1/2 + 1/3 * 1 + 0} = 1/3 = 33\% $

You incorrectly wrote $P(D=3 \mid H=2) = 1$ That is incorrect.
$$P(D=3\mid H=2) = \frac{P(D=3,H=2)}{P(H=2)} = \frac{1/6}{1/3} = \frac{1}{2}$$
Make that correction and you have:
$P(H=1|D=3) = p(H=1)  \frac{p(D=3|H=1}{\sum_{i=1} ^3 p(H=i) * p(D=3|H=i)} = \frac{1/3 * 1/2} { 1/3 * 1/2 + 1/3 * 1/2 + 0} = 1/2$
which is correct
An additional comment:
This analysis does not solve the Monty Hall problem because it completely neglects the door $C$ that the contestant chooses.
