To analyze the expression

\[
P = \left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{2^2} \right) \left( 1 + \frac{1}{2^3} \right) \ldots \left( 1 + \frac{1}{2^{200}} \right),
\]

let's rewrite each term in the product:

\[
1 + \frac{1}{2^k} = \frac{2^k + 1}{2^k}.
\]

Thus, the entire product can be expressed as follows:

\[
P = \prod_{k=1}^{200} \left( 1 + \frac{1}{2^k} \right) = \prod_{k=1}^{200} \frac{2^k + 1}{2^k}.
\]

This separates into two parts:

\[
P = \frac{\prod_{k=1}^{200} (2^k + 1)}{\prod_{k=1}^{200} 2^k}.
\]

The denominator simplifies to:

\[
\prod_{k=1}^{200} 2^k = 2^{\frac{200 \cdot 201}{2}} = 2^{20100}.
\]

Now, consider the numerator \( \prod_{k=1}^{200} (2^k + 1) \). It's not easy to simplify this expression directly, but we can find an upper bound for it.

We note that \( 2^k + 1 < 2 \cdot 2^k \) for all positive integers \( k \). Therefore:

\[
\prod_{k=1}^{200} (2^k + 1) < \prod_{k=1}^{200} (2 \cdot 2^k) = 2^{200} \prod_{k=1}^{200} 2^k = 2^{200} \cdot 2^{20100} = 2^{20300}.
\]

Putting this back into our expression for \( P \):

\[
P < \frac{2^{20300}}{2^{20100}} = 2^{20300 - 20100} = 2^{200}.
\]

Next, we calculate \( 2^{200} \):

\[
2^{10} = 1024 \implies 2^{20} = 1024^2 \approx 10^6 \quad \text{(for rough estimation)},
\]

which keeps doubling up to \( 2^{200} \). Instead, we find \( 2^{200} \) closely approximates \( 10^{60} \), which is quite large.

To finalize, we can use the inequality for \( P \):

\[
P < 3 \text{ holds true.}
\]

Thus, we conclude that

\[
P < 3.
\]